Logic and Proofs
Jeff Edmonds
York University
Lecture 1
Propositional Logic
And, Or, Not, Implies
Predicates and Quantifiers
boys b, Loves(b)
Quick Examples
Notation Truth Tables
What This Logic Is and Is Not
Venn Diagrams
Tautologies, Equivs, Proofs
Arguing Our Rules/Lemmas
• And vs Or
• Implies
• Deduction
• Cases
• Contradiction
• Distributive
• Review
Example Proofs
• Tautologies, Equivs, Proofs
• Axiom Schema (Darwin)
• Prove No Racism
Objects, Predicates, & Relations
&
Order of Quantifiers: vs
Quantifiers and Games
Humans are Mortal
Reals
Formal Proofs Systems
(More EECS1090):
Review
Things cut out
(More EECS1090):
Puzzle
1. All snakes are reptiles
2. Some reptiles hatch their eggs
themselves
Animals
Reptiles
Snakes Hatch
Themselves
Can we conclude:
Some snakes hatch their eggs themselves? Y/N
No: It could be all the reptiles that hatch themselves
are not snakes.
A single counter example suffices.
Puzzle
1. All snakes are reptiles
2. All reptiles are animals and lay eggs
3. Many snakes only have one lung
Animals
Lay Eggs Reptiles
Snakes
Can we conclude:
Some animals that lay eggs only have one lung.
Yes:
Proof by example/picture does not suffice.
One lung
Puzzle
1. All snakes are reptiles
2. All reptiles are animals and lay eggs
3. Many snakes only have one lung
Animals
Lay Eggs Reptiles
Snakes
One lung
Similar to many questions in course/degree/life
• Read question in English
• Assign variables to each
atomic true/false logical proposition/statement/assertion.
• Translate what you know
and the question into logic.
• Solve the logic using logic rules,
formal, and informal proofs
• Translate solution back to English.
Puzzle
1. All snakes are reptiles
2. All reptiles are animals and lay eggs
3. Many snakes only have one lung
Animals
Lay Eggs Reptiles
Snakes
X One lung
exists
Goal: x, animal(x) & eggs(x) & onelung(x)
forall
1. ∀x, snake(x) reptile(x)
2. ∀x, reptile(x) (animal(x) & eggs(x))
3. x, snake(x) & onelung(x)
4. Let X be such that snake(X) & onelung(X)
5. snake(X)
6. reptile(X)
7. animal(X) & eggs(X)
8. animal(X) & eggs(X) & onelung(X)
9. x, animal(x) & eggs(x) & onelung(x)
Some animals that lay eggs only have one lung.
Given
Given
Given
By 3
By 4
By 1 & 5
By 2 & 6
By 5 & 7
By 8
In English
Puzzle
1. All snakes are reptiles
2. All reptiles are animals and lay eggs
3. Many snakes only have one lung
Animals
Lay Eggs Reptiles
Snakes
One lung
Similar to many questions in course/degree/life
• Read question in English
• Assign variables to each
atomic true/false logical proposition/statement/assertion.
• Translate what you know
and the question into logic.
• Solve the logic using logic rules,
formal, and informal proofs
• Translate solution back to English.
Quick Examples
You can have
(ham and eggs)or porridge
You can have
coffee and(eggs or porridge)
Aaaaaah!
Restaurants: exclusive or
Logic:
inclusive or
Quick Examples
True/False Variables:
L1 “Mrs Peacock loves Colonel Mustard”
L2 “Colonel Mustard loves Miss Scarlet”
K “it happened in the kitchen”
P “Mrs Peacock committed the murder”.
Quick Examples
If
L1 “Mrs Peacock loves Colonel Mustard”
and L “Colonel Mustard loves Miss Scarlet”
2
and K “it happened in the kitchen”
then
P “Mrs Peacock committed the murder”.
L1and
L2 and
KP
If all on left are true,
then that on right is true.
Quick Examples
If
L1 “Mrs Peacock loves Colonel Mustard”
and L “Colonel Mustard loves Miss Scarlet”
2
and K “it happened in the kitchen”
then
P “Mrs Peacock committed the murder”.
L1and
L2 and
KP
P
and and
or
or
K
(L1 L2 K ) ≡ L1 L2
These can’t all
At least one of these
be true.
must be false
Notation
Proof System Notation:
Variables (true/false)
• P, Q
Formulas (true/false)
• α, β:
α and β are true.
• αβ:
is shaped like A for And.
α or β are true (or both) a bucket to put anything
• αβ:
Not α.
• α
• αβ: If α, then β.
hound dog
• α iff β: α if and only β.
hound dog
Exclusive or (not both). Parity.
• αβ:
“Meta” Notation: ie not within the proof system.
• αβ: If α is a line in the proof, then β can be the next line.
• α⊢β: If α is an axiom, then β can be proved. i.e. αα1α2…β
• α⊧β:
If α, then β. (αβ states our universe having has property.
α⊧β states true in every universe.)
Notation
Instead of
Φ ≡ x y ⌐(αβ) γδωµπσ
Γ = {Φ, }
Feel free to write
Forall x exits y -(a or b) --> g & d & w & u & p & s
Multiplication and Addition Tables
Sometimes,
we evaluate
a formula given
a specific assignment.
Evaluate with
x=2 and y=3.
(x+y)y
(2+3)3
5 3
15
x
y
(x+y)y
2
3
15
Multiplication and Addition Tables
(x+y)y
(2+4)4
6 4
24
Evaluate with
x=2 and y=4.
x
y
(x+y)y
2
3
15
4
24
2
Multiplication and Addition Tables
Sometimes,
(x+y)y
we want to know
≡
how it does on ALL
(xy) + (yy)
∞ assignments.
Distributive
Sometimes,
Law
we use this is our
reasoning about life.
x
y
(x+y)y
2
3
15
This
24
2 4Equivalence
…
is a Tautology/Valid
(ie always true)
Operator's Truth Tables
Q
Q
and
T
P
F
T
T
F
F
F
F
P
Sometimes,
we evaluate
a formula given
a specific assignment.
Evaluate with
A=T and B=F.
Sometimes,
we want to know
how it does on 2n
ALL assignments.
or
T
F
T
T
T
F
T
F
or
Q
not
P
implies
T
F
F
T
P
T
F
T
T
F
F
T
T
and
(AB) (AB)
(TF) (TF)
T F
F
A
B
T
T
T
F
F
T
F
F
(AB) (AB)
F
Operator's Truth Tables
Q
Q
and
T
P
F
T
T
F
F
F
F
P
or
T
F
T
T
T
F
T
F
or
Evaluate with
A=T and B=T.
not
P
implies
T
F
F
T
P
T
F
T
T
F
F
T
T
and
(AB) (AB)
(TT) (TT)
Knowing
T thatT
at leastTone is true,
does not imply
that both are true.
Sometimes,
we want to know
how it does on
ALL 2n assignments.
Q
A
B
(AB) (AB)
T
T
T
T
F
F
F
T
F
F
F
T
Sometimes,
we use this is our
reasoning about life.
Operator's Truth Tables
Q
Q
and
T
P
F
T
T
F
F
F
F
P
or
T
F
T
T
T
F
T
F
or
not
P
Evaluate with
A=T and B=F.
implies
T
F
F
T
P
T
F
T
T
F
F
T
T
and
(AB) (AB)
and
Q
A
B
(AB) (AB)
T
T
T
T
F
F
F
T
F
F
F
T
A
B
(AB) (AB)
T
T
T
F
F
T
F
F
or
(AB) (AB)
(TF) (TF)
F T
T
T
Operator's Truth Tables
Q
Q
and
T
P
F
T
T
F
F
F
F
P
or
T
F
T
T
T
F
T
F
or
not
P
implies
T
F
F
T
P
T
F
T
T
F
F
T
T
and
(AB) (AB)
and
Q
A
B
(AB) (AB)
T
T
T
T
F
F
F
T
F
F
F
T
A
B
(AB) (AB)
T
T
T
or
(AB) (AB)
Knowing that
both are true,
implies that
at least one is true.
T
F
F
ThisT is a
T
Tautology/Valid
T
T true)
F always
(ie
F
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T T
T
T
F
T
T
F
T F
F
T
T
F
F
T
F T
F
T
T
F
T
F
F
F
T
T
F
F
Connectors: , , , , iff, , …
• And (PQ): P and Q are both true.
– It is raining and I am wet
Q
P
T
F
T
T
F
F
F
F
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T T
T
T
F
T
T
F
T F
F
T
T
F
F
T
F T
F
T
T
F
T
F
F
F
T
T
F
F
Connectors: , , , , iff, , …
• And (PQ…R): All of P, Q, and R are true.
– It is raining and I am wet and I am outside
– We don’t add brackets because ((PQ)R) = (P(QR))
Q
P
T
F
T
T
F
F
F
F
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T T
T
T
F
T
T
F
T F
F
T
T
F
F
T
F T
F
T
T
F
T
F
F
F
T
T
F
F
Connectors: , , , , iff, , …
• And (PQ…R): All of P, Q, and R are true.
• Or (PQ…R): At least one of P, Q, and R are true.
Q
P
T
F
T
F
T
T
F
T
T
T
F
F
F
F
T
F
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T T
T
T
F
T
T
F
T F
F
T
T
F
F
T
F T
F
T
T
F
T
F
F
F
T
T
F
F
Connectors: , , , , iff, , …
• And (PQ…R): All of P, Q, and R are true.
• Or (PQ…R): At least one of P, Q, and R are true.
• Negation (P): Has the flipped true/false value.
P
T
F
T
F
T
T
F
T
T
T
T
F
F
F
F
F
T
F
F
T
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
TT
T
T
F
T
T
F
TF
F
T
T
F
F
T
F T
F
T
T
F
T
F
F
F
T
T
F
F
Connectors: , , , , iff, , …
• And (PQ…R): All of P, Q, and R are true.
• Or (PQ…R): At least one of P, Q, and R are true.
• Negation (P): Has the flipped true/false value.
• Implies (PQ): If P is true then so is Q.
Q
P
T
F
T
F
T
T
F
T
T
T
T
F
F
F
F
T
F
F
T
F
F
T
T
F
T
F
F
T
Possibilities:
• P being true causes Q to be true
• Q being true causes P to be true
• R being true causes P and Q to be true
• Impossible for P and Q.
• Coincident
In statistics, the mantra is
“Correlation Causation”
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T T
T
T
F
T
T
F
T F
F
T
T
F
F
T
F T
F
T
T
F
T
F
F
F
T
T
F
F
Connectors: , , , , iff, , …
• And (PQ…R): All of P, Q, and R are true.
• Or (PQ…R): At least one of P, Q, and R are true.
• Negation (P): Has the flipped true/false value.
• Implies (PQ): If P is true then so is Q.
– If P is false, then (PQ) is true no matter what Q is.
Q
“hound” “dog”
P
T
F
T
F
T
T
F
T
T
T
T
F
F
F
F
T
F
F
T
F
F
T
T
F
T
F
T
T
is true even if “hound”.
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T=T
T
T
F
T
T
F
T F
F
T
T
F
F
T
F T
F
T
T
F
T
F=F
F
F
T
T
F
Connectors: , , , , iff, , …
• And (PQ…R): All of P, Q, and R are true.
• Or (PQ…R): At least one of P, Q, and R are true.
• Negation (P): Has the flipped true/false value.
• Implies (PQ): If P is true then so is Q.
• If and only if (P iff Q): P and Q are both true or both false.
Q
P
T
F
T
F
T
T
F
T
T
T
T
F
F
F
F
T
F
F
T
F
iff
T
F
F
T
T
F
T
T
F
T
F
T
T
F
F
T
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T T
T
T
F
T
T
F
T F
F
T
T
F
F
T
F T
F
T
T
F
T
F
F
F
T
T
F
F
Connectors: , , , , iff, , …
• Exclusive-Or (PQ): One has be true but not both.
You can have eggs or porridge
But not both!
Q
P
T
F
T
F
T
T
F
T
T
T
T
F
F
F
F
T
F
F
T
F
iff
T
F
T
F
F
T
T
F
T
T
F
T
F
T
T
F
T
T
F
F
T
F
T
F
Operator's Truth Tables
And
Or
P Q PQ PQ
T=T
T F
All
combinations
F T
F=F
2
1
1
0
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T
T
F
T
T
F
F
T
T
F
F
T
F
T
T
F
T
F
F
T
T
F
Connectors: , , , , iff, , …
• Exclusive-Or (PQ): One has be true but not both.
P and Q have the opposite values.
– Parity (PQ…R): The number of true variables is odd.
Q
P
T
F
T
F
T
T
F
T
T
T
T
F
F
F
F
T
F
F
T
F
iff
T
F
T
F
F
T
T
F
T
T
F
T
F
T
T
F
T
T
F
F
T
F
T
F
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T T
T
T
F
T
T
F
T F
F
T
T
F
F
T
F T
F
T
T
F
T
F
F
F
T
T
F
F
Connectors: , , , , iff, , …
• Exclusive-Or (PQ): One has be true but not both.
P and Q have the opposite values.
– Parity (PQ…R): The number of true variables is odd.
This is because if you flip the value of any of these variables,
the result always changes.
We love this function because it is sensitive to each variable.
P
T
F
T
F
T
T
F
T
T
T
T
F
F
F
F
T
F
F
T
F
iff
T
F
T
F
F
T
T
F
T
T
F
T
F
T
T
F
T
T
F
F
T
F
T
F
Operator's Truth Tables
And
Or
P Q PQ PQ
All
combinations
Not
Implies
If and
only if
Exclusive
Or
Q
PQ
P iff Q
PQ
T T
T
T
F
T
T
0
T F
F
T
T
F
F
1
F T
F
T
T
F
1
F
F
F
T
T
0
F
Connectors: , , , , iff, , …
• Exclusive-Or (PQ): One has be true but not both.
P and Q have the opposite values.
– Parity (PQ…R): The number of true variables is odd.
– Boolean Addition: 1+0+1+1+0 = 3 =mod2 1
Translate truth to 1 and false to 0.
The # of truth is 1+0+1+1+0 = 3
T F
+mod2 1
Taken mod2 is the 1
T F T
1 0
# is odd.
F T F
0 1
0
1
0
Electronic circuits
Values:
• 0 ≈ False ≈ off ≈ 0 volts
• 1 ≈ True ≈ on ≈ 5 volts
Gates:
Can compute any Boolean function:
Boolean strings:
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
Eg: “All people are created equal”
“We are all equal, just some of us are more equal than others.”
I like this as a statement
about inequity
But even more as a joke
about logic.
In Propositional Logic, each formula/variable evaluates to true/false.
These are combined with
Not, And, Or, Implies, and Iff
Sentences are equivalent
, , , ,
iff
≡
32
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
In contrast, Predicate Logic (next topic)
talks about the properties of objects.
Parent(x,y) ≡ “x is y’s parent”
Daughter(y,x) iff Parent(x,y)Female(y)
33
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α≡
You may talk about
But this has no place
in this class!
β≡
α is either true or false
or independent
of what we know.
Not knowing is cause by
not knowing which
universe/model/
interpretation/assignment
we are in.
In each of these, α is either
true or false.
34
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α≡
You may talk about
β≡
α is either true or false
or independent
of what we know.
There is no gray.
But this has no place
in this class!
35
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α≡
You may talk about
β≡
α is either true or false
or independent
of what we know.
Hoping will not help.
But this has no place
in this class!
36
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α≡
You may talk about
β≡
α is either true or false
or independent
of what we know.
We won’t consider
different people with
different beliefs.
And we are never wrong.
But this has no place
in this class!
37
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α≡
You may talk about
β≡
α is either true or false
or independent
of what we know.
There are no probability
distributions.
And hence no “likely”?
But this has no place
in this class!
38
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α≡
You may talk about
But this has no place
in this class!
β≡
α is either true or false
or independent
of what we know.
There is no time.
The answer
does not change.
And hence no
“now” or “later”.
39
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α≡
You may talk about
β≡
It is not possible that
and
β is false and αβ is true.
and
This is by definition of .
When you don’t know
you can say
“maybe”.
40
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α≡
You may talk about
β≡
αβ means:
“If is α true, then so is β.”
But …..
there is no sense of causality.
AAAAAH !
This is the confusing one!
But this has no place
in this class!
41
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
β
α
α
β
or
αβ
T
F
T
T
T
T
T
T
T
F
T
F
T
F
F
T
T
F
F
F
Just like the table that
states 4+3=7,
we have table for αβ.
αβ means:
“If is α true, then so is β.”
But …..
there is no sense of causality.
42
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
α
β αβ
T
T
T
T
F
F
F
T
F
F
β
T
α
F
Just like the table that
states 4+3=7,
we have table for αβ.
And for αβ.
αβ means:
“If is α true, then so is β.”
But …..
there is no sense of causality.
43
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
β
T
α
T
F
T
F
α
β αβ
T
T
T
F
F
T
F
F
T
If α=β=T, then αβ
is clearly true.
This is the example,
αβ was meant for.
αβ means:
“If is α true, then so is β.”
But …..
there is no sense of causality.
44
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
β
α
α
β αβ
T
F
T
T
T
T
F
T
F
F
F
T
F
F
F
T
If α=T and β=F, then αβ
is clearly false.
This is another example,
αβ was meant for.
αβ means:
“If is α true, then so is β.”
But …..
there is no sense of causality.
45
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
β
α
α
β αβ
and
αβ
T
F
T
T
T
T
T
F
T
F
F
F
T
?
F
T
T
?
F
F
F
T
?
F
F
T
T
?
If α=F, then αβ is ???.
We don’t know about β.
The statement αβ is not useful.
But let’s make αβ true.
This table is the definition of .
If the bottom two are F,
Then this operation would be And.
αβ means:
“If is α true, then so is β.”
But …..
there is no sense of causality.
46
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
β
α
α
β αβ
Hound
Dog
HoundDog
T
F
T
T
T
T
T
T
T
T
F
T
F
F
T
F
F
F
T
T
F
T
T
F
T
T
F
F
T
F
F
T
If α=F, then αβ is ???.
We don’t know about β.
The statement αβ is not useful.
But let’s make αβ true.
This table is the definition of .
αβ means:
“If is α true, then so is β.”
But …..
there is no sense of causality.
No problem!
That makes sense.
47
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
β
α
α
β αβ
Elephant
Pink ElephantPink
T
F
T
T
T
T
T
T
T
T
F
T
F
F
T
F
F
F
T
T
F
T
T
F
T
T
F
F
T
F
F
T
If α=F, then αβ is ???.
We don’t know about β.
But Let’s make αβ true.
This table is the definition of .
It is all about the table.
If Elephant=Pink=F,
then ElephantPink is True.
αβ means:
“If is α true, then so is β.”
But …..
there is no sense of causality.
AAAAAH !
This is the confusing one!
48
What Our Logic Is and Is Not
Max( A )
“preCond:
Input is array A[1..n]
of n values.”
i=1
m = A[1]
loop
“loop-invariant:
m is max in {A[1]..A[i]}”
exit when (i=n)
m = max(m,A[i+1])
i=i+1
endloop
return(m)
“postCond:
return max in {A[1]..A[n]}”
My boss wants me to write code.
He gives me:
The contract is
preCond postCond
Preconditions:
Any assumptions that must be true
about the input instance.
Postconditions:
Any statement of what must be true
when the algorithm/program
returns.
I give him:
49
What Our Logic Is and Is Not
Max( A )
“preCond:
Input is array A[1..n]
of n values.”
i=1
m = A[1]
loop
“loop-invariant:
m is max in {A[1]..A[i]}”
exit when (i=n)
m = max(m,A[i+1])
i=i+1
endloop
return(m)
“postCond:
return max in {A[1]..A[n]}”
My boss wants me to write code.
He gives me:
The contract is
preCond postCond
T
T
If the client gives us an input that
meets the precondition
and my program gives an output that
meets the postcondition,
then everyone is happy.
50
What Our Logic Is and Is Not
Max( A )
“preCond:
Input is array A[1..n]
of n values.”
i=1
m = A[1]
loop
“loop-invariant:
m is max in {A[1]..A[i]}”
exit when (i=n)
m = max(m,A[i+1])
i=i+1
endloop
return(m)
“postCond:
return max in {A[1]..A[n]}”
My boss wants me to write code.
He gives me:
The contract is
preCond postCond
F
F
If the client gives us an input that
does NOT meet the precondition
and my program gives an output that
does NOT meet the postcondition,
then my boss can’t be mad.
My code meets the contract.
51
What Our Logic Is and Is Not
Max( A )
My boss wants me to write code.
“preCond:
He gives me:
Input is array A[1..n]
The contract is
of n values.”
preCond postCond
i=1
T
F
m = A[1]
The only way to break my contract is if
loop
the input that meets the precondition
“loop-invariant:
and my output does NOT meet the postcondition.
m is max in {A[1]..A[i]}”
exit when (i=n)
m = max(m,A[i+1])
i=i+1
endloop
return(m)
“postCond:
return max in {A[1]..A[n]}”
52
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
• An Assignment will tell us
the True/False value of all our variables
• A Model M will tell us
if we are dealing with reals or integers or people.
Together we will call this a Universe.
There are many possibilities.
53
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
But I don’t
know what
is true.
In a given
universe
Each sentence is either
true or false
not both
54
What Our Logic Is and Is Not
In English, a Proposition is a statement or assertion.
But I don’t
know what
is true.
That is
because you
do not know
which
universe we
are in.
55
What Our Logic Is and Is Not
α β
α is true or false
β is true or false
This gives 4
possibilities.
α,β =
T,F
T,T
αβ
T
T
T
T
F
F
F
T
T
F
F
T
F,T F,F
56
What Our Logic Is and Is Not
α β
In which is αβ true?
“If is α true,
then so is β”
can be confusing
because people think
of causality.
α,β =
T,F
T,T
αβ
T
T
T
T
F
F
F
T
T
F
F
T
F,T F,F
57
What Our Logic Is and Is Not
α β
Suppose I tell you αβ.
α,β =
F,T
T,F
Instead, use the
T,T
operation’s table.
“It is not true that
F
α is true β is not.”
αβ = T
αβ
T
T
T
T
F
F
F
T
T
F
F
T
F,F
T
T
58
What Our Logic Is and Is Not
α β
Telling you α is true,
eliminates other
possibilities.
α,β =
T,F
T,T
αβ
T
T
T
T
F
F
F
T
T
F
F
T
F,T F,F
T
T
F
αβ = T
59
What Our Logic Is and Is Not
α β
In all the worlds
remaining possible,
β is true.
α,β =
T,F
T,T
αβ
T
T
T
T
F
F
F
T
T
F
F
T
F,T F,F
T
T
F
αβ = T
60
What Our Logic Is and Is Not
Modus Ponens:
From α and αβ, conclude β.
α β
“If is α true,
then so is β”
Hey we still do use this!
From this we can
conclude ….
α,β =
T,F
T,T
αβ
T
T
T
T
F
F
F
T
T
F
F
T
F,T F,F
T
T
F
αβ = T
61
What Our Logic Is and Is Not
Modus Ponens:
From α and αβ, conclude β.
α β
This sentence is
said to be a
tautology/valid
because it is true
in every universe.
α,β =
T,F
T,T
αβ
T
T
T
T
F
F
F
T
T
F
F
T
F,T F,F
T
T
F
αβ = T
62
What Our Logic Is and Is Not
Modus Ponens:
From α and αβ, conclude β.
α β
This sentence is
said to be a
tautalogy/valid
because it is true
in every universe.
α,β =
T,F
T,T
αβ
T
T
T
T
F
F
F
T
T
F
F
T
F,T F,F
T
T
F
αβ = T
63
Different Ways of Expressing p→q
if p, then q
if p, q
p implies q
p is sufficient for q
q if p
q whenever p
q follows from p
q when p
p only if q
q is necessary for p
???
Says ¬q→¬p
which is the same as p→q
Converses of each other.
64
64
Different Ways of Expressing p→q
if p, then q
if p, q
p implies q
p is sufficient for q
a sufficient condition for q is p
q if p
q whenever p
q follows from p
q when p
p only if q
q is necessary for p
a necessary condition for p is q
65
65
Different Ways of Expressing p↔q
(p→q) and (q→p).
p if and only if q
p iff q
p is necessary and sufficient for q.
p implies q and the converse
and conversely.
The converse of p→q is q→p.
66
66
Dude! You will be starting with things that are
A good martial arts student will attentively repeat
each fundamental technique many times.
Don’t tune out when a concept
is taught more than once.
The better students listen carefully in order to
refine and develop their understanding.
67
Venn Diagrams
α
α=T
β=F
β
α=T α=F
β=T β=T
(αβ) = F (αβ) = T (αβ) = T
α = F, β = F, (αβ) = T
α β
αβ
T
T
T
T
F
F
F
T
T
F
F
T
Let this denote
the set of all
universes/possibilities/people/objects/….
Let this denote
those in which α is true.
Let this denote
those in which is β true.
This gives the same four possibilities
as in the table.
68
α
Venn Diagrams
α
β
β
and
αβ
α
α
β
β
Set intersection
ie only if both
For every object,
Obj in set αβ
iff in α and in β.
Otherwise,
falls out of cup .
For every universe,
αβ is true
iff α and β are true.
Otherwise,
falls out of cup .
69
α
Venn Diagrams
α
β
β
αβ
β
For every universe,
αβ is true
iff α or β are true.
Stays in cup .
For every object,
Obj in set αβ
iff in α or in β.
Stays in cup .
or
α
α
β
Set union
ie if either
70
¬α
Venn Diagrams
α
¬β
β
and
¬α¬β
Which area is
green in both iff ¬(αβ)
Which
area is
pictures
green in at
above?
least one
or
picture?
¬α¬β
iff ¬(αβ)
De Morgan's Law
α
α
β
β
Set intersection
ie only if both
α
β
Set union
ie if either
71
Venn Diagrams
and
and
α¬β
α
β
¬(α¬β)
α
β
α
β
and
¬(α¬β)
In addition to knowing ¬(α¬β), iff αβ
suppose we also know α.
We can conclude β.
For every object,
obj in set α
iff in β.
α
This could be the set
of universes I
which I am a hound,
Or the set of hounds.
subset
hound dog
hound dog
72
Venn Diagrams
and
and
α¬β
α
β
¬(α¬β)
α
β
and
¬(α¬β)
α
In addition to knowing ¬(α¬β), iff αβ
α
iff βα
suppose we also know α.
iff ¬β¬α
We can conclude β.
Contrapositive
In addition to knowing ¬(α¬β),
suppose we also know ¬β.
We can conclude ¬α.
β
subset
hound dog
hound dog
73
αβ
iff ¬β¬α
Contrapositive
• American Dream
≡ “If you are a good person, then you will be rich.
≡ “If you are not rich, then you are not a good person.
74
or
¬αβ
Venn Diagrams
and
α
β
¬(α¬β)
β
α
and
¬(α¬β)
In addition to knowing ¬(α¬β), iff αβ
iff βα
suppose we also know α.
iff ¬β¬α
We can conclude β.
or
iff ¬αβ
DeMorgan
α
β
subset
hound dog
hound dog
75
Venn Diagrams
In addition to knowing αβ,
suppose we also know βγ.
We can conclude αγ.
Transitivity
and
¬(α¬β)
iff αβ
iff βα
iff ¬β¬α
or
iff ¬αβ
γ
α
β
subset
hound dog
hound dog
76
For every object,
Venn
Diagrams
it is in both or in neither
In addition to knowing αβ,
in α iif in β.
suppose we also know βα.
We can conclude β iff α.
For every universe,
both are true or neigher
α is true iff β is true.
Equivalence
βα
α
αβ
β
α
β
and
subset
β iff α
α
β
subset
hound dog
hound dog
equal sets
77
Tautologies, Equivalences, & Proofs
• Tautology: A proposition that is true under every assignment.
or
Example: p¬p.
• Contradiction: A proposition that is false under every assignment.
and
Example: p¬p
• Satisfiable: A propositions that is true under at least one assignment.
p
T
F
¬p
F
T
p∨¬p
T
T
p∧¬p
F
F
78
78
Tautologies, Equivalences, & Proofs
• Equivalent, p≡q: Two propositions that have same true/false
under every assignment.
p↔q is a tautology.
2# variables cells!!!
> # atoms in the universe.
p
T
T
q
T
F
¬p
F
F
F
F
T
F
T
T
Equivalent
Not Equivalent
q→p
¬p ∨ q p → q
T
F
T
F
T
T
T
T
T
T
F
T
79
79
Tautologies, Equivalences, & Proofs
A Hilbert Style Proof is a sequence of formulas.
Hilbert 1920
1.
2.
3.
4.
5.
6.
7.
n.
1
2
3
4
5
6
….
• Each line is either:
– i Axiomslogical Axiomsnon-logical
Logical Axioms: Axiomslogical
• A set of formulas that has already
been proved to always be true.
Non-Logical Axioms Γ:
• A set of formulas that have been given by the
problem to tbe problem true.
80
Tautologies, Equivalences, & Proofs
A Hilbert Style Proof is a sequence of formulas.
Hilbert 1920
1.
2.
3.
4.
5.
6.
7.
1
2
3
4
5
6
n.
….
• Each line is either:
– i Axiomslogical Axiomsnon-logical
– Follows from a lemma/rule of the form:
“If i and j are lines of your proof,
then you can add k as a line of your proof.”
– Each such line k should be documented
where it came from.
• The last line of the proof is our theorem .
81
Tautologies, Equivalences, & Proofs
Eval/Build
Separating And
αβ
→ α
Selecting Or
αβ & α
→
Eval/Build
β
Cases
αβ, αγ, & βγ → γ
Modus Ponens
α & αβ
β
→
Cases
αβ, αγ, & βγ →
αβ
→
→ (βγ)
α&β
β
γ
α
α & β
→
αγ
(αβ)
Excluded Middle
αα & (αα)
Deduction
α→β
Assume α, prove β
Eval/Build
β
α
α & β
γβ
→ αγ
→ (αβ)
→
In every possible universe,
if the Left-Hand-Side
Transitivityis true,
then so
is &
theβγ
Right-Hand-Side.
αβ
→ αγ
These are very
→
By Contradiction: From α(β&β),
α.
importantInconsistent:
and we will
detail later.
From study
β and β,them inanything.
→
82
Tautologies, Equivalences, & Proofs
Here are some very useful equivalences
Equivalence
α→β & β→α iff α iff β
Double Negation: α iff α
Commutative: αβ iff βα and αβ iff βα
Distributive: γ(αβ) iff (γα)(γβ) and γ(αβ) iff (γα)(γβ)
Contrapositive
αβ iff βα iff ¬αβ
De Morgan's Law
¬(αβ) iff ¬α¬β
83
Tautologies, Equivalences, & Proofs
Eval/Build
Separating And
αβ
→ α
Selecting Or
αβ & α
→
Eval/Build
β
Cases
αβ, αγ, & βγ → γ
Modus Ponens
α & αβ
β
→
Cases
αβ, αγ, & βγ →
αβ
→
→ (βγ)
α&β
β
γ
Equivalence
α→β & β→α iff α iff β
Transitivity
αβ & βγ → αγ
α
α & β
→
αγ
(αβ)
Excluded Middle
αα & (αα)
Deduction
α→β
Assume α, prove β
Eval/Build
β
α
α & β
γβ
→ αγ
→ (αβ)
→
Contrapositive
αβ iff βα iff ¬αβ
De Morgan's Law
¬(αβ) iff ¬α¬β
→
By Contradiction: From α(β&β),
α.
Double Negation: α iff α
anything.
Commutative: αβ iff βα and αβ iff βα Inconsistent: From β and β,
Distributive: γ(αβ) iff (γα)(γβ) and γ(αβ) iff (γα)(γβ)
→
84
And
Or
Proof Techniques/Lemmas
Using:
Proving:
From:
From:
Conclude:
Conclude:
Eval/Build
Separating And
α&β
αβ
αβ
α&β
β
(βγ)
Tautologies, Equivalences, & Proofs
And :
Or :
Selecting Or
αβ & α
Cases
αβ, αγ, & βγ
Implies
:
Modus Ponens
α & αβ
Cases
αβ, αγ, & βγ
Eval/Build
β
γ
β
γ
Equivalence
α→β & β→α
α iff β
Transitivity
αβ & βγ
αγ
α
α & β
αγ
(αβ)
Excluded Middle
αα & (αα)
Deduction
α→β
Assume α, prove β
Eval/Build
β
α
α & β
γβ
αγ
(αβ)
Contrapositive
αβ iff βα iff ¬αβ
De Morgan's Law
¬(αβ) iff ¬α¬β
By Contradiction: From α(β&β), conclude α.
Double Negation: α iff α
Commutative: αβ iff βα and αβ iff βα Inconsistent: From β and β, conclude anything.
Distributive: γ(αβ) iff (γα)(γβ) and γ(αβ) iff (γα)(γβ)
85
And
Or
Proof Techniques/Lemmas
Using:
Proving:
From:
From:
Conclude:
Conclude:
Eval/Build
Separating And
α&β
αβ
αβ
α&β
β
(βγ)
Tautologies, Equivalences, & Proofs
And :
Or :
Selecting Or
αβ & α
Cases
αβ, αγ, & βγ
Implies
:
Modus Ponens
α & αβ
Cases
αβ, αγ, & βγ
Eval/Build
β
γ
β
γ
α
α & β
αγ
(αβ)
Excluded Middle
αα & (αα)
Deduction
α→β
Assume α, prove β
Eval/Build
β
α
α & β
γβ
αγ
(αβ)
Each lemma/rule has the form:
Equivalence
Contrapositive
“Ifα→β
i and
are
lines
of
your
proof,
j
& β→α
α iff β
αβ iff βα iff ¬αβ
then youTransitivity
can add k as a line of your proof.”
αβ & βγ
αγ
De Morgan's Law
¬(αβ) iff ¬α¬β
By Contradiction: From α(β&β), conclude α.
Double Negation: α iff α
Commutative: αβ iff βα and αβ iff βα Inconsistent: From β and β, conclude anything.
Distributive: γ(αβ) iff (γα)(γβ) and γ(αβ) iff (γα)(γβ)
86
Arguing Our Rules/Lemmas
Proofs of this date back
thousands of years.
4 (½ ab) + c2 =
(a+b)2
= a2 + 2ab + b2
a2 + b2 = c2
Babylon and Egypt
Pythagoras
(400 BC)
(1900 B.C.).
“Let there be proofs”
And it was good.
Euclid (300 BC)
A Proof System:
• Is a mechanical method of proving
that some formula is true.
87
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
“I am doing logic”and
“I love logic”
αβ
α
β
Non-Logical Axiom
There is not much you can PROVE to be true
without starting with some assumptions.
Such assumptions are called Non-Logical Axioms.
We don’t prove them.
But our conclusion is only as strong as they are.
88
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
“I am doing logic”and
“I love logic”
αβ
α
β
Non-Logical Axiom
At each step in the proof,
you can add any line that follows from the previous lines.
There is an implied “implies”
{ previous lines} next line.
89
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1. “I am doing logic”and
“I love logic”
2. “I am doing logic”
αβ
α
β
α
Non-Logical Axiom
What follows from this single line?
Give me a weaker statement
ie one that is guaranteed to be true
within every universe in which the first is true.
Exactly!
90
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
αβ
α
β
α
“I am doing logic”and
“I love logic”
“I am doing logic”
Non-Logical Axiom
Separating And: 1
It is good to use established rules in your proof.
and
Separating And: From αβ, conclude α.
• If both are true, then each must be true individually.
It is good to label each line with the rule used.
That was fun. Let’s do it again.
91
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
3.
“I am doing logic”and
“I love logic”
“I am doing logic”
or
“I am doing logic” “I love logic”
αβ
α
αβ
α
β
Non-Logical Axiom
Separating And: 1
Our next line must follow from a combination of previous.
Give a new line that is a weakening of the second.
ie true within every universe in which the second is true.
92
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
3.
“I am doing logic”and
“I love logic”
“I am doing logic”
or
“I am doing logic” “I love logic”
αβ
α
αβ
α
β
Non-Logical Axiom
Separating And: 1
Evaluating/Building: 2
Our next line must follow from a combination of previous.
Give a new line that is a weakening of the second.
ie true within every universe in which the second is true.
Evaluating/Building:
or
• From α alone, conclude αγ.
If I know α=T, I can evaluate αγ to be true. From α, I can build αγ.93
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
3.
and
“I love logic”
“I am doing logic” or
“I am doing logic” or
and
“I love logic”
“I am doing logic”
αβ
αβ
α
αβ
αβ
α
β
Non-Logical Axiom
Separating And: 1
Evaluating/Building: 2
Could we turn this around?
Are these legal proof steps?
Oops. No.
In the first, we only know at least one is true.
We do not know which.
The other’s require knowing that you are doing logic.
94
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
F
T
αβ
T α
α F
αβ F
F
β
1.
2.
3.
T
F
“I am doing logic” or
“I love logic”
F logic”
“I am doing
and T
F
“I am doing logic” “I love logic”
T
F
F
Non-Logical Axiom
XXXX
XXXX
Could we turn this around?
Are these legal proof steps?
The universe in which I love it, but don’t do it
has the first true and others false.
Hence, the second two do not follow from the first.
95
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
T
F
αβ
T α
α T
αβ F
T
β
1.
2.
3.
F
T
“I am doing logic” or
“I love logic”
T logic”
“I am doing
and F
T
“I am doing logic” “I love logic”
T
T
F
Non-Logical Axiom
XXXX
XXXX
Could we turn this around?
Are these legal proof steps?
The universe in which I do it, but don’t love it
has the first two true and third false.
Hence, the third does not follow from the first two.
96
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
αβ
or
“I am doing logic” “I love logic”
α
β
Non-Logical Axiom
If each line of the proof is weaker,
how do we prove strong statements?
We have our new lines follow
from two or more previous lines.
97
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
3.
αβ
& β
α
or
“I am doing logic” “I love logic”
“I love logic”
“I am doing logic”
α
β
Non-Logical Axiom
Non-Logical Axiom
Selecting Or: 1 & 2
What follows from these two lines?
Exactly!
Selecting Or: From αβ and β, conclude α.
• If at least one of α and β is true,
and β is not,
then α must be.
98
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
And
α β αβ
1.
2.
3.
or
“I am doing logic” “I love logic”
“I love logic”
“I am doing logic”
Or
αβ
T T
T
T
T F
F
T
F T
F
T
F
F
F
F
99
Arguing Our Rules/Lemmas
And vs Or
α ≡ “I am doing logic”
β ≡ “I love logic”
And
α β αβ
1.
2.
3.
or
“I am doing logic” “I love logic”
“I love logic”
“I am doing logic”
And
Or
And :
Or :
Or
αβ
T T
T
T
T F
F
T
F T
F
T
F
F
F
F
Proof Techniques/Lemmas
Using:
Proving:
From:
From:
Conclude:
Conclude:
Eval/Build
Separating And
α&β
αβ
αβ
α&β
α
(αγ)
Selecting Or
αβ & α
Eval/Build
β
α
α & β
αγ
(αβ)
100
Arguing Our Rules/Lemmas
Implies
α ≡ “I am doing logic”
β ≡ “I love logic”
If
1.
2.
3.
α β
αβ
T
T
T
T
F
F
F
T
T
F F
then
“I love logic” “I am doing logic”
“I love logic”
“I am doing logic”
T
αβ
&α
β
α
α
β
Non-Logical Axiom
Non-Logical Axiom
Modus Ponens: 1 & 2
What follows from these two lines?
Exactly!
Modus Ponens: From α and αβ, conclude β.
• If α is true and α ensures β, then β must be true.
101
Arguing Our Rules/Lemmas
Implies
α ≡ “I am doing logic”
β ≡ “I love logic”
If
1.
2.
α β
αβ
T
T
T
T
F
F
F
T
T
F F
T
then
F
F
“I love logic” “I am doing logic” T
F logic”
F
“I love
F
F
αβ
T
α
αF F
β
Non-Logical Axiom
Non-Logical
Axiom
XXXX
????
Does the second line follow from the first?
Oops. No.
It says “IF I love logic”.
It does not say that I love it.
The universe in which I do it, but don’t love it
has the first true and second false.
102
Arguing Our Rules/Lemmas
Implies
α ≡ “I am doing logic”
β ≡ “I love logic”
If
1.
2.
α β
αβ
T
T
T
T
F
F
F
T
T
F
T
αβ
T
α
T
F
βα F
F F
T
then
T
F
“I love logic” “I am doing logic” T
F
T
“I am doing logic” “I love logic” F
β
Non-Logical Axiom
XXXX
What follows from this single line?
Oops. No.
hounddog does not imply doghound
The universe in which I do it, but don’t love it
has the first true and second false.
Hence, the second does not follow from the first.
103
Arguing Our Rules/Lemmas
Implies
α ≡ “I am doing logic”
β ≡ “I love logic”
αβ
βα
1.
2.
“I love logic” “I am doing logic”
“I am doing logic” “I love logic”
α
β
Non-Logical Axiom
Contrapositive
What follows from this single line?
Exactly!
Contrapositive: From αβ, conclude βα.
• If α ensures β and β is not true, then α can’t be.
Otherwise β would be.
104
Arguing Our Rules/Lemmas
Implies
α ≡ “I am doing logic”
β ≡ “I love logic”
αβ
βα
1.
2.
“I love logic” “I am doing logic”
“I am doing logic” “I love logic”
α
β
Non-Logical Axiom
Contrapositive
If you want to know the truth,
I will tell you.
Dude!
You will tell me
whether I want to you to or not.
When β=T,
γβ is vacuously true.
105
Arguing Our Rules/Lemmas
Implies
α ≡ “I am doing logic”
β ≡ “I love logic”
αβ
βα
1.
2.
“I love logic” “I am doing logic”
“I am doing logic” “I love logic”
α
β
Non-Logical Axiom
Contrapositive
There will be no racism,
when hell freezes over.
Dude!
Hell, by definition, Hell will never freeze over.
Hence, you are implying
that there will always be racism.
Minimally, you are saying a vacuous statement.
When α=F,
αγ is vacuously true.
106
Arguing Our Rules/Lemmas
Implies
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
α β
αβ
T
T
T
T
F
F
F
T
T
F
F
T
“I love logic” “I am doing logic”
“I am doing logic” “I love logic”
Non-Logical Axiom
The nextContrapositive
rules will
define this table.
Proof Techniques/Lemmas
Proving:
From:
From:
Conclude:
Conclude:
Eval/Build
β
Modus Ponens
γβ
α
αγ
α & αβ
β
Using:
Implies
:
α & β
Equivalence
α→β & β→α
α iff β
Transitivity
αβ & βγ
αγ
(αβ)
Contrapositive
¬αβ iff βα iff αβ
Transitivity
αβ & βγ
αγ
107
Evaluate/Build
And
More
complex
things.
Or
P Q PQ PQ
T T
T
T
Not
Implies
Q
PQ
F
T
T F
F
T
F T
F
T
T
F
F
F
T
F
T
F
Only
F
Let’splug
understand
the following concepts
into
andplaces
see the equivalence between them.
needed.
Evaluate Formula
From value α=F,
Evaluate [(αβ) (βγ)] [αγ]
?
?
[F?]
T
T
Proof Techniques/Lemmas
Proving:
From:
Conclude:
Eval/Build
α&β
β
αβ
(βγ)
Eval/Build
α
α & β
αγ
(αβ)
Eval/Build
β
α
α & β
γβ
αγ
(αβ)
Build Formulas in Proofs
1.
2.
3.
α
Proved before
αγ
Eval/Build : 1
[(αβ) (βγ)] [αγ]
Eval/Build : 2
108
Evaluate/Build
Sometimes
the proof
forms a tree.
And
Or
P Q PQ PQ
T T
T
T
Not
Implies
Q
PQ
F
T
T F
F
T
F T
F
T
T
F
F
F
T
F
T
(Tβ) (Fγ)
T
T
T
α&β
β
αβ
(βγ)
Eval/Build
α
α & β
F
αγ
(αβ)
Eval/Build
β
α
α & β
Let’s understand the following concepts
and see the equivalence between them.
Evaluate Formula
From value α=T,
Evaluate (αβ) (αγ)
Proof Techniques/Lemmas
Proving:
From:
Conclude:
Eval/Build
γβ
αγ
(αβ)
Build Formulas in Proofs
1.
2.
3.
4.
α
αβ
αγ
(αβ) (αγ)
Proved before
Eval/Build : 1
Eval/Build : 1
Eval/Build : 1
109
Arguing Our Rules/Lemmas
Deduction
α ≡ “I am doing logic”
β ≡ “I love logic”
From:
Implies
:
α
Using:
Proof Techniques/Lemmas
Conclude:
Modus Ponens
α & αβ
β
From:
β
Proving:
Conclude:
Deduction
α→β
Assume α, prove β
110
Arguing Our Rules/Lemmas
Deduction
α ≡ “I am doing logic”
β ≡ “I love logic”
β
α
Deduction: Assume α, prove β, conclude α→β.
If α is false, then α→β is automatically true.
We temporally assume α is true,
and with this we need to prove β is true.
α β
αβ
T
T
T
T
F
F
F
T
T
F
F
T
111
Arguing Our Rules/Lemmas
Deduction
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
3.
4.
5.
Deduction Goal: “I love logic” “I am doing logic”
“I love logic”
“If I love x” “I do x”
“I am doing logic”
“I love logic” “I am doing logic”
α
β
Assumption
Non-Logical Axiom
Modus Ponens: 2 & 3
Deduction Conclusion
Deduction Format:
Deduction Goal: αβ
It is indented
______ α
Assumption
to remind
______ …
within this scope
______ β
Proved somehow
αβ
Deduction Conclusion is α assumed.
112
Arguing Our Rules/Lemmas
Cases
The set of all people
α ≡ “I am doing logic”
β ≡ “I love logic”
α
I want to prove
“I love logic”.
There are two cases.
Do you love it in the day?
Do you love it in the night?
β
γ
If α is true, then so is γ.
If β is true, then γ.
Either α or β is true.
Hence, γ is true.
Cases Goal: γ.
Cases α & β.
1. Given:
2.
α or β, ie, these cover all the cases
3.
Case 1: Assume α and prove γ.
4.
Case 2: Assume β and prove γ.
5.
Case 3: Assume ???
6. Conclude γ.
113
Arguing Our Rules/Lemmas
Cases
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
3.
4.
5.
Cases Goal: “I love logic”.
“Daytime” or “Nighttime”
“Daytime” “I love logic”
“Nighttime” “I love logic”
“I love logic”
Cases “Daytime” or “Nighttime”
Proved somehow.
Proved somehow.
Proved somehow.
Cases 2, 3, 4
Cases Goal: γ.
Cases α & β.
1. Given:
2.
α or β, ie, these cover all the cases
3.
Case 1: Assume α and prove γ.
4.
Case 2: Assume β and prove γ.
This is a rule just
5. Conclude γ.
like the others.
114
Arguing Our Rules/Lemmas
Cases
α ≡ “I am doing logic”
β ≡ “I love logic”
1.
2.
3.
4.
5.
Cases Goal: “I love logic”.
“Daytime” or “Nighttime”
“Daytime” “I love logic”
“Nighttime” “I love logic”
“I love logic”
From:
Implies
:
Cases “Daytime” or “Nighttime”
Proved somehow.
Proved somehow.
Proved somehow.
Cases 2, 3, 4
Proof Techniques/Lemmas
Using:
Proving:
From:
Conclude:
Conclude:
Cases
αβ, αγ, & βγ
γ
Deduction
α→β
Assume α, prove β
It feels natural to merge
cases and deduction
into one process..
115
Arguing Our Rules/Lemmas
Cases
α ≡ “I am doing logic”
β ≡ “I love logic”
Cases Goal: γ.
1. αβ
2. Case 1:
3.
α
4.
….
5.
γ
6. Case 2:
7.
β
8.
….
9.
γ
10. γ
Or tighten it up a bit.
Cases α & β
Proved somehow
Assumption
Proved somehow
Assumption
Proved somehow
Cases Conclusion 1, 5, 9
116
Arguing Our Rules/Lemmas
Cases
α ≡ “I am doing logic”
β ≡ “I love logic”
Cases Goal: γ.
1. αβ
2. Case α:
3.
4.
….
5.
γ
6. Case β:
7.
8.
….
9.
γ
10. γ
Or tighten it up a bit.
Cases α & β
Proved somehow
Assumption
Proved somehow
Assumption
Proved somehow
Cases Conclusion 1, 5, 9
117
Arguing Our Rules/Lemmas
Cases
α ≡ “I am doing logic”
β ≡ “I love logic”
Cases Goal: γ.
1. α1α2…αn
2. Case α1:
3.
…
4. Case α2:
5.
…
6. Case α3:
7.
8.
…
9. Case αn:
10. γ
Or tighten it up a bit.
Cases α1, α2, …, & αn
Proved somehow
Assumption
Assumption
Assumption
Assumption
Cases Conclusion
118
Arguing Our Rules/Lemmas
Cases
1.
2.
3.
4.
5.
6.
7.
8.
9.
Cases Goal: “I love logic”.
“Daytime” or “Nighttime”
Case “Daytime”:
…
“I love logic”
Case 2: “Nighttime”
…
“I love logic”
“I love logic”
Cases “Daytime” or “Nighttime”
Proved somehow
Assumption
Proved somehow
Assumption
Proved somehow
Cases Conclusion
119
Arguing Our Rules/Lemmas
Cases
1.
2.
3.
4.
5.
6.
7.
8.
9.
Cases Goal: “Everyone is rich”.
“America” or “Europe”
Case “America”:
…
“Everyone is rich”
Case “Europe”:
…
“Everyone is rich”
“Everyone is rich”
Cases “America” or “Europe”
Proved somehow
Assumption
Proved somehow
Assumption
Proved somehow
Cases Conclusion
What???
There are more cases!
120
Arguing Our Rules/Lemmas
Contradiction
Contradiction: ββ.
• In our logic, β can’t be both true and false.
This is called a contradiction.
121
Arguing Our Rules/Lemmas
Contradiction
By Contradiction: From α(β&β), conclude α.
This is a classic technique for proving α.
• “By way of contradiction assume the opposite”, ie α.
• From it, prove a contradiction.
• Hence, the initial assumption must be false, ie is α true.
Proof of Proof By Contradiction:
1.
2.
3.
4.
α (β&β)
(β&β) α
(β&β)
α
Proved somehow
Contra Positive from 1
Excluded middle
Modus Pontus from 2&4
122
Arguing Our Rules/Lemmas
Contradiction
The Greeks only knew about integers and fractions.
These were called rational.
They did not know about decimals x = 1.41421356237…
But they could not express √2 as a fraction.
But how do you prove that a fraction representation
DOES NOT EXIST?
If we assumed the opposite,
then we have √2 = n/d to work with.
123
Arguing Our Rules/Lemmas
Proof that √2 is irrational.Contradiction
Let α be “√2 is irrational”.
Let β be “the number of 2 divisors in 2d2=n2 is even”
Goal αβ
By way of contradiction assume α, i.e “√2 is rational”.
By definition of rational, √2 = n/d for integers n&d.
Rearrange 2d2=n2
Every integer can be uniquely factored into primes.
Eg 12=223
Repeatedly divide n by 2 until the remaining value is odd.
This gives n=2kr for some int k and some odd r.
n2=22kr2, giving the number of 2 divisors in it is even.
Hence β.
Hence αβ
124
Arguing Our Rules/Lemmas
Proof that √2 is irrational.Contradiction
Let α be “√2 is irrational”.
Let β be “the number of 2 divisors in 2d2=n2 is even”
Goal αβ
By way of contradiction assume α, i.e “√2 is rational”.
By definition of rational, √2 = n/d for integers n&d.
Rearrange 2d2=n2
Every integer can be uniquely factored into primes.
Eg 12=223
Repeatedly divide d by 2 until the remaining value is odd.
This gives d=2k'r' for some int k and some odd r.
2d2 =22k'+1r'2, giving the num of 2 divisors in it is odd.
Hence β.
Hence αβ
125
Arguing Our Rules/Lemmas
Proof that √2 is irrational.Contradiction
Let α be “√2 is irrational”.
Let β be “n&d are not both even”
We proved αβ and αβ
Hence α(β&β)
Because α proves a contradiction, α must be wrong.
Hence, α must be true.
Ie “√2 is irrational”.
126
Arguing Our Rules/Lemmas
Contradiction
The Greeks knew about primes 2, 3, 5 ,7, …
There seems to be a lot of them.
But how do you prove that there are
an INFINITE number of them.
If we assumed the opposite,
then we have a maximum to work with.
127
Arguing Our Rules/Lemmas
Contradiction
Proof that there are an infinite number of primes.
Let α be “there are an infinite number of primes”.
Let β be “p is the maximum prime”
Goal αβ
By way of contradiction assume α, i.e “finite number”.
Hence, there must be a maximum prime p.
Hence β.
Hence αβ
128
Arguing Our Rules/Lemmas
Contradiction
Proof that there are an infinite number of primes.
Let α be “there are an infinite number of primes”.
Let β be “p is the maximum prime”
Goal αβ
Let n = p!+1
= p(p-1)(p-2)…1 + 1
Every integer can be factored into primes, eg 12 = 223.
What are the prime factors of n?
Is p a factor of n?
No, the remainder of n/p is 1.
Is p-1 a factor of n?
No, the remainder of n/(p-1) is 1.
Nothing p or less is a factor of n.
129
Arguing Our Rules/Lemmas
Contradiction
Proof that there are an infinite number of primes.
Let α be “there are an infinite number of primes”.
Let β be “p is the maximum prime”
Goal αβ
Hence, n must have a prime factor bigger than p.
Hence, there is a prime bigger than p.
Hence β.
Hence αβ
130
Arguing Our Rules/Lemmas
Contradiction
Proof that there are an infinite number of primes.
Let α be “there are an infinite number of primes”.
Let β be “p is the maximum prime”
We proved αβ and αβ
Hence α(β&β)
Because α proves a contradiction, α must be wrong.
Hence, α must be true.
Ie “there are an infinite number of primes”.
Note we gave an algorithm for constructing
bigger and bigger and bigger primes.
Never ending.
131
Arguing Our Rules/Lemmas
Contradiction
By Contradiction: From α(β&β), conclude α.
This is a classic technique for proving α.
• “By way of contradiction assume the opposite”, ie α.
• From it, prove a contradiction.
• Hence, the initial assumption must be false, ie is α true.
It is useful when
assuming the contradiction
allows you to assume
the existence of some object.
But this technique is way over used.
Please avoid it by proving directly.
132
Arguing Our Rules/Lemmas
Distributive
A Tautology is a formula that evaluates to true
under every setting of the variables.
Proof: A tautology can be proved by
checking every setting of the variables.
•
Distributive: x(y+z) = (xy)+(xz)
Area
Area
xy
y+z x(y+z)
z xz
x
y
133
Arguing Our Rules/Lemmas
Distributive
A Tautology is a formula that evaluates to true
under every setting of the variables.
Proof: A tautology can be proved by
checking every setting of the variables.
•
Distributive: x(y+z)
x+(yz)
2+(11)
3
= (xy)+(xz)
= (x+y)(x+z)
= (2+1)(2+1)
=
9
134
Arguing Our Rules/Lemmas
Distributive
A Tautology is a formula that evaluates to true
under every setting of the variables.
Proof: A tautology can be proved by
checking every setting of the variables.
•
Distributive: x(y+z)
x+(yz)
P(QR)
P(QR)
= (xy)+(xz)
= (x+y)(x+z)
iff (PQ)(PR)
iff (PQ)(PR)
P
Q
R
P(QR)
(PQ)(PR)
P
Q
R
P(QR)
(PQ)(PR)
T
T
T
T
T
T
T
T
T
T
T
T
F
T
T
T
T
F
T
T
T
F
T
T
T
T
F
T
T
T
T
F
F
F
F
T
F
F
T
T
F
T
T
F
F
F
T
T
T
T
F
T
F
F
F
F
T
F
T
T
F
F
T
F
F
F
F
T
T
T
F
F
F
F
F
F
F
F
F
F
135
Arguing Our Rules/Lemmas
Distributive
A Tautology is a formula that evaluates to true
under every setting of the variables.
Proof: A tautology can be proved by
checking every setting of the variables.
•
Distributive: x(y+z) = (xy)+(xz)
x+(yz) = (x+y)(x+z)
P(QR) iff (PQ)(PR)
Between P and others is .
136
Arguing Our Rules/Lemmas
Distributive
A Tautology is a formula that evaluates to true
under every setting of the variables.
Proof: A tautology can be proved by
checking every setting of the variables.
•
Distributive: x(y+z) = (xy)+(xz)
x+(yz) = (x+y)(x+z)
P(QR) iff (PQ)(PR)
Between Q and R is .
137
Rules/Lemmas
The goal is to proveArguing Our
Distributive
Distributive: γ(αβ) iff (γα)(γβ)
γ(αβ) iff (γα)(γβ)
γ
(αβ)
α
β
γ
α
β
=
γ
(γα)
γ(αβ)
α
β
(γβ)
α
β
γ
α
γ
(γα)(γβ)
α
β
β
γ
=
(αβ)
α
β
γ
α
(γα)
α
γ
=
(γβ)
β
α
γ
γ
(γα)(γβ)
α
β
β
γ
Compare areas.
Same
γ(αβ)
α
β
β
γ
γ
=
γ
Compare areas.
Same
138
Without Loss of Generality
Generality means that something is true in General,
i.e. in all situations. i.e. a Tautology.
In proof by cases: You must prove ALL of the cases.
But if two of the cases are the same except that something
symmetric has been stitched, then you can say:
“Without Loss of Generality
we will consider the first case and ignore the second.”
Prove: min(x, y) = (x + y − |x − y|)∕2
Here x and y are symmetric, so we can say
“Without Loss of Generality assume x − y ≥ 0, then
(x + y − |x − y|)∕2 = (x + y − x + y|)∕2 = (2y)∕2 = y
= min(x, y)
139
And
Or
And :
Or :
Proof Techniques/Lemmas
Using:
Proving:
From:
From:
Conclude:
Conclude:
Eval/Build
Separating And
α&β
αβ
αβ
α&β
β
(βγ)
Selecting Or
αβ & α
Cases
αβ, αγ, & βγ
Implies
:
Modus Ponens
α & αβ
Cases
αβ, αγ, & βγ
Eval/Build
β
γ
β
γ
Equivalence
α→β & β→α
α iff β
Transitivity
αβ & βγ
αγ
α
α & β
αγ
(αβ)
Excluded Middle
αα & (αα)
Deduction
α→β
Assume α, prove β
Eval/Build
β
α
α & β
γβ
αγ
(αβ)
Contrapositive
αβ iff βα iff ¬αβ
De Morgan's Law
¬(αβ) iff ¬α¬β
By Contradiction: From α(β&β), conclude α.
Double Negation: α iff α
Commutative: αβ iff βα and αβ iff βα Inconsistent: From β and β, conclude anything.
Distributive: γ(αβ) iff (γα)(γβ) and γ(αβ) iff (γα)(γβ)
140
141
142
143
144
145
146
Tautologies, Equivalences, & Proofs
• Equivalent, p≡q: Two math formulas that have same real value
under every assignment.
(x+y)2 – x2 - y2 + 2x
= (x2 + 2xy + y2) – x2 - y2 + 2x
= 2xy + 2x
= 2x (y+1)
147
Tautologies, Equivalences, & Proofs
• Equivalent, p≡q: Two propositions that have same true/false
under every assignment.
p
T
T
q
T
F
¬p
F
F
F
F
T
F
T
T
Equivalent
Not Equivalent
q→p
¬p ∨ q p → q
T
F
T
F
T
T
T
T
T
T
F
T
148
148
Tautologies, Equivalences, & Proofs
• Equivalent, p≡q: propositions have same true/false
under every assignment.
Equivalence
α→β & β→α
α iff β
Double Negation: α iff α
Commutative: αβ iff βα and αβ iff βα
Distributive: γ(αβ) iff (γα)(γβ) and γ(αβ) iff (γα)(γβ)
Contrapositive
αβ iff βα iff ¬αβ
De Morgan's Law
¬(αβ) iff ¬α¬β
149
Key Logical Equivalences
1
Identity Laws:
p T p,
pF p
Domination
Laws:
Idempotent
laws:
Double Negation
Law:
Negation Laws:
pT T,
pF F
p p p,
pp p
Absorption
Laws:
© McGraw Hill LLC
p p
p p T ,
p p F
p p q p p p q p
150
More Logical Equivalences
TABLE 7 Logical Equivalences
Involving Conditional
Statements.
p q p q
p q q p
p q p q
p q p q
p q p q
p q p r p q r
p r q r p q r
p q p r p q r
p r q r p q r
© McGraw Hill LLC
TABLE 8 Logical Equivalences
Involving Biconditional
Statements.
p q p q q p
p q p q
p q p q p q
p q p q
151
Tautologies, Equivalences, & Proofs
p p q
is logically equivalent to p q
Example: Show that
Solution:
p p q p p q
p p q
p p q
by the second De Morgan law
by the first De Morgan
law
by the double negation law
p p p q
by the second distributive law
F p q
because
p q F
p q
p p F
by the commutative law
for disjunction
By the identity law for F
152
152
Tautologies, Equivalences, & Proofs
Example: Show that
p q p q
is a tautology.
Solution:
p q p q p q p q
p q p q
p p q q
by truth table for →
by the first De Morgan
law
by associative and
commutative laws
laws for disjunction
T T
by truth tables
T
by the domination law
153
153
Tautologies, Equivalences, & Proofs
• Disjunctive Normal Form (DNF): Or of Ands
and
or
(p¬q) (r¬st) ¬q
• Conjunctive Normal Form (CNF): And of Os
and
(p¬q) (r¬st) ¬q
or
• Every formula can be put uniquely into DNF (CNF)
(pq)r (¬p¬q) ¬r
No clause is the subset of another.
or
(pq) (pqr)
and
and
154
Tautologies, Equivalences, & Proofs
• Disjunctive Normal Form (DNF): Or of Ands
and
or
(p¬q) (r¬st) ¬q
• Conjunctive Normal Form (CNF): And of Os
and
(p¬q) (r¬st) ¬q
or
• Every formula can be put uniquely into DNF (CNF)
(pq)r (¬p¬q) ¬r
• One proof that two formula are equivalent is
to put both into DNF and then compare.
Time ≥ # clauses
≥ 2# variables
≥ # atoms in the universe.
eg pqr…t
155
Tautologies, Equivalences, & Proofs
• Disjunctive Normal Form (DNF): Or of Ands
and
or
(p¬q) (r¬st) ¬q
• Conjunctive Normal Form (CNF): And of Os
and
(p¬q) (r¬st) ¬q
or
• Finding a satisfying assignment for CNF is hard,
because for each clause must choose one.
But this is hard because you must be consistent with T/F values.
• Finding a satisfying assignment for DNF is easy,
because only one clause needs to be satisfied.
• Finding a falsifying assignment for DNF is hard.
Our goal is to find faster
ways to prove things.
156
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
“dolphins nurse young” “dolphins are mammals”
“dolphins are mammals” “dolphins are warm blooded”
“dolphins nurse young”
I would like to generalize to animals
I have never seen before.
Proof
Axioms
1. “dolphins nurse young” “dolphins are mammals”
2. “dolphins are mammals” “dolphins are warm blooded”
3. “dolphins nurse young”
Modus Ponens: 1&3
4. “dolphins are mammals”
Modus Ponens: 2&4
5. “dolphins are warm blooded”
157
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
“x nurse young” “x are mammals”
“x are mammals” “x are warm blooded”
“dolphins nurse young”
Any objects x can be plugged in.
I would like to generalize to animals
I have never seen before.
An axiom schema is a formula in the metalanguage of an axiomatic
system, in which one or more schematic variables appear. These
variables, which are metalinguistic constructs, stand for any term or
subformula of the system, which may or may not be required to
satisfy certain conditions.
158
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
“x nurse young” “x are mammals”
“x are mammals” “x are warm blooded”
“dolphins nurse young”
Any objects x can be plugged in.
I would like to generalize to animals
I have never seen before.
Meta-proof
1. “x nurse young” “x are mammals”
2. “x are mammals” “x are warm blooded”
3. “x nurse young” “x are warm blooded”
Axiom
Axiom
Transitivity
159
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
4.
“x nurse young” “x are mammals”
“x are mammals” “x are warm blooded”
“dolphins nurse young”
“humans nurse young”
Learning is fun:
1. “humans nurse young”
2. “humans nurse young” “humans are mammals”
3. “humans are mammals”
4. “humans are mammals” “humans are warm blooded”
5. “humans are warm blooded”
Axiom
Axiom
Modus Ponens: 1&2
Axiom
Modus Ponens: 3&4
160
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
4.
“x nurse young” “x are mammals”
“x are mammals” “x are warm blooded”
“dolphins nurse young”
“humans nurse young”
But these are not true for lizards!
α
β αβ
T
T
T
T
F
F
F
T
T
F
F
T
Yes, they are!
161
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
4.
5.
“x nurse young” “x are mammals”
“x are mammals” “x are warm blooded”
“dolphins nurse young”
“humans nurse young”
“lizards are warm blooded”
Learning is fun:
1. “lizards are warm blooded”
2. “lizards are mammals” “lizards are warm blooded”
3. “lizards are warm blooded” “lizards are mammals”
4. “lizards are mammals”
5. “lizards nurse young”
Axiom
Axiom
Contra Positive: 2
Modus Ponens: 1&3
And so on …
162
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
4.
5.
“x nurse young” “x are mammals”
“x are mammals” “x are warm blooded”
“dolphins nurse young”
“humans nurse young”
hound dog
“lizards nurse young”
dog hound
dog hound
hound dog
Learning is fun:
1. “lizards nurse young”
2. “lizards nurse young” “lizards are mammals”
3. Oops we are stuck!
Axiom
Contra Positive
Converse
Axiom
Axiom
163
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
4.
5.
iff“x are mammals”
“x nurse young”
“x are mammals” “x are warm blooded”
“dolphins nurse young”
“humans nurse young”
hound dog
“lizards nurse young”
dog hound
dog hound
hound dog
Learning is fun:
1. “lizards nurse young”
2. “lizards nurse young” “lizards are mammals”
3. Oops we are stuck!
4. “lizards are mammals” “lizards nurse young”
5. “lizards nurse young” “lizards are mammals”
6. “lizards are mammals”
Axiom
Contra Positive
Converse
Axiom
Axiom
Axiom
Contra Positive: 4
Modus Ponens: 1&5
164
Axiom Schema
Non-Logical Axioms Γ:
1.
2.
3.
4.
5.
6.
iff “x are mammals”
“x nurse young”
“x are mammals”iff??“x are warm blooded”
“dolphins nurse young”
“humans nurse young”
“birds are mammals”
“birds are warm blooded”
Learning is fun:
165
Example: Prove No Racism
I will prove
“I will get a million dollars”
“a world without racism”
166
Example: Prove No Racism
α ≡ “I fail” β ≡ “My mom is unhappy”
167
Example: Prove No Racism
1.
2.
“I fail” “My mom is unhappy”
[“I fail” “My mom is unhappy”] “I fail”
Clear
Parse Tree
“I fail”
“My mom is unhappy”
“I fail”
What does this mean???
In this line 2, what is the most important connector?
Root of parse tree ie the last one that is applied.
What is the Left and Right-HandSide of this ?
168
Example: Prove No Racism
1.
2.
“I fail” “My mom is unhappy”
[“I fail” “My mom is unhappy”] “I fail”
Clear
Clear
What does this mean???
The line says “If Left-Hand-Side, then Right-HandSide”.
Oh yes:
If my mother would be unhappy if I fail,
then I will be sure not to fail!
169
Example: Prove No Racism
1.
2.
3.
“I fail” “My mom is unhappy”
[“I fail” “My mom is unhappy”] “I fail”
α
β
“I fail”
Clear
Clear
How do we use this?
Modus Ponens: From α and αβ, conclude β.
• If α is true and α ensures β, then β must be true.
Look we already know the LHS, ie α.
Hence, we can conclude the RHS, ie β.
170
Example: Prove No Racism
1.
2.
3.
4.
5.
“I fail” “My mom is unhappy”
[“I fail” “My mom is unhappy”] “I fail”
“I fail”
α
“I study” “I fail” β
“I fail” “I study”
Clear
Clear
Modus Ponens 2&3
Clear
Contrapositive
Contrapositive: From αβ, conclude βα.
• If α ensures β and β is not true, then α can’t be.
Otherwise β would be.
171
Example: Prove No Racism
1.
2.
3.
4.
5.
6.
7.
“I fail” “My mom is unhappy”
[“I fail” “My mom is unhappy”] “I fail”
“I fail”
“I study” “I fail”
“I fail” “I study”
“I study”
“I study”
Clear
Clear
Modus Ponens 2&3
Clear
Contrapositive
Modus Ponens 3&5
Double Negation
Double Negation: From α, conclude α.
• In our logic, something is either true or false.
The middle is excluded.
If something is not (not true), then it must be true.
172
Example: Prove No Racism
1.
2.
3.
4.
5.
6.
7.
“I study”
Remember this. We will need it later.
173
Example: Prove No Racism
1.
2.
3.
4.
“I am sick” “I study”
β
α
“I am sick” “I party”
β “I study”γ
“I party”
“I am sick” “I study”
Clear
Clear
Clear
Transitivity 2&3
Transitivity: From αβ and βγ, conclude αγ.
• If I can travel from α to β and from β to γ
then I can travel from α to γ.
174
Example: Prove No Racism
1.
“I am sick” “I study”
2.
3.
4.
“I am sick” “I study”
5.
or
“I am sick” “I am sick”
Excluded Middle
or
Excluded Middle: ββ.
• In our logic, something can’t be neither true nor false.
The universe must decide.
175
Example: Prove No Racism
1.
2.
3.
4.
γ
α
“I am sick” “I study”
5.
γ
β
“I am sick” “I study”
β
αor
“I am sick” “I am sick”
6.
“I study”
By Cases 5,1,4
or
By Cases: From αβ, αγ, and βγ, conclude γ.
1. Given: There are only two cases α & β.
2.
Case 1: Assume α and prove γ.
3.
Case 2: Assume β and prove γ.
4. Conclude γ.
176
Example: Prove No Racism
1.
2.
3.
“I study”
“I study”
“I study”and
“I study”
Proved on 1st page
Proved on 2nd page
Build/Evaluate 1&2
Evaluating/Building:
and
• From both α and α´, conclude αα´.
177
Example: Prove No Racism
1.
2.
3.
“I study”
“I study”
“I study”and
“I study”
4.
“a world without racism”
5. qed
Latin for quod "Which was to be demonstrated."
This does not make any sense.
Try again!
Inconsistent: From β and β, conclude anything.
• A funny twist of our logic is that
once you have proved a contradiction,
from it you can conclude anything you want.
178
Example: Prove No Racism
1.
(“I study” “I study”)
2.
(“I study” “I study”)
“a world without racism”
“I study” “I study”
“a world without racism”
3.
4.
Excluded Middle
Can’t be both true and false.
Evaluate/Build : 1
From α, conclude αanything
Proved on previous page of proof
Modus Ponens 2&3
179
Example: Prove No Racism
This proof must have it’s flaws.
One of these assumptions was probably wrong:
•
[“I fail” “My mom is unhappy”] “I fail”
•
“I am sick” “I party”
180
Example: Prove No Racism
Because they are illogical.
No
Such assumptions are called Non-Logical Axioms:
•
[“I fail” “My mom is unhappy”] “I fail”
We don’t prove them.
But we choose to assume that they are true.
181
Example: Prove No Racism
Because they are illogical.
No
Such assumptions are called Non-Logical Axioms:
•
[“I fail” “My mom is unhappy”] “I fail”
We don’t prove them.
But we choose to assume that they are true.
182
Predicates and Quantifiers
boys b, Loves(b)
183
Objects, Predicates, & Relations
Predicates: boy(x), father(x), ...
is True if and only if object x has stated property.
Relations: loves(b,g), fatherOf(x,y), ...
is True if and only if objects have stated relation.
Defining:
fatherOf(x,y)exist
≡ “x is y’s father”.
father(x) ≡ y fatherOf(x,y)
daughter(y,x) iff father(x,y)girl(y)
184
Objects, Predicates, & Relations
Predicate logic can only say these stupid things.
Why should I care?
Actually, it is hugely expressive.
• Math: For mathematical statement that you want to understand
and prove, it is best to first write it as a Predicate logic sentence.
• Proof: Sometimes the mechanics of the proof just falls out.
• All Computation: There is a Predicate logic sentence over the integers (+,)
Compute(J,I,y) ≡ “J(I)=y”, for every Java program J,
ie Compute(J,I,y) is true iff J(I) outputs y.
• Halting Problem: There is another Halt(J,I) ≡ “J(I) halts”,
which is not itself computable.
• Hard: The bad news is that this means that there is no algorithm
determining whether such a sentence is true.
185
Objects, Predicates,
& & Relationsg
Loves(b,g)
People need to
love and be loved.
Sam
Mary
Bob
Beth
T
T
John
Marilyn
Monroe
T
T
Fred
Ann
b
T
T
T
T
Loves(Sam,Mary) = False
Loves(Sam,Beth) = True
“There
exists a boy that loves Mary”
exist
b, Loves(b,Mary)
False
b, Loves(b,Beth)
True
Sorry
“Every boy loves Beth”
this is heteronormative.
all
b, Loves(b,Beth)
False
b, Loves(b,MM)
186
True
&
g, [b, Loves(b, g)]
b, [g, Loves(b, g)]
loop g{girls} (checking true for some)
loop b{boys} (checking true for all)
Loves(b,g)
loop b{boys} (checking true for all)
loop g{girls} (checking true for some)
Loves(b,g)
g
g
T
T
b
T
T
T
b
T
T
T
187
g, [b, Loves(b, g)]
b
g
T
T
T
T
T
b
T
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
Loves(b,g)
g
g
g
g
g
g
g
T
T
b
And
Or
Or
Or
Or
Or
Or
And And And And And
b
b
b
b
b
b, [g, Loves(b, g)]
&
188
vs
In all three examples,
every boy loves a girl.
The difference?
Sam
Mary
Bob
Fred
Beth
Marilyn
Monroe
Ann
Sam
Mary
Bob
Beth
Marilyn
Monroe
Ann
John
One girl
John
Fred
T
T
T
T
T
T
T
T
T
T
T
T
T
Could be a separate girl.
Sam
Mary
Bob
Beth
Marilyn
Monroe
Ann
John
Fred
His special woman.
T
T
T
T
189
vs
“There is a girl
for all boys to love.”
“There is a inverse (eg 1/3)
for all reals.”
Not clear.
Sam
Mary
Bob
Fred
Beth
Marilyn
Monroe
Ann
Sam
Mary
Bob
Beth
Marilyn
Monroe
Ann
John
One girl
John
Fred
T
T
T
T
T
T
T
T
T
T
T
T
T
Could be a separate girl.
“For each boy,
there is a girl.”
“For each person,
there is God.”
Sam
Mary
Bob
Beth
Marilyn
Monroe
Ann
John
Fred
His special woman.
T
T
T
T
190
vs
This statement is “about” girls.
The existence of one that is
“loved by everyone”.
b,[g, Loves(b, g) ]
Mary
Bob
Fred
Beth
Marilyn
Monroe
Ann
Sam
Mary
Bob
Beth
Marilyn
Monroe
Ann
John
One girl
g,[b, Loves(b, g)]
Sam
John
Fred
T
T
T
T
T
T
T
T
T
T
T
T
T
Could be a separate girl.
This statement is “about” boys.
Each of them
“loves someone.”
Sam
Mary
Bob
Beth
Marilyn
Monroe
Ann
John
Fred
His special woman.
T
T
T
T
191
vs
This creates a function p=p(v).
Sam
Fred
p
Beth
Ann
b, g, Loves(b, g)
Sam
Mary
Bob
Beth
Marilyn
Monroe
Ann
John
Fred
His special woman.
T
T
T
T
192
vs
Sam
Mary
Bob
Beth
Marilyn
Monroe
Ann
John
One girl
Fred
T
T
T
T
A:g, b, Loves(b, g)
B:b, g, Loves(b, g)
Prove A B
If there is one girl that is loved by every boy,
Then every boy loves at least that girl.
193
vs
“There is a politician that is loved by everyone.”
Sam
Bob
John
Trudeau
Fred
politician, voters, Loves(v, p)
voters, politician, Loves(v, p)
Sam
“Every voter loves some politician.”
Bob
John
Scheer
Trudeau
Fred
194
vs
x, y, y=f(x)
This only says
that f is a function
that is defined on
the whole domain.
y, x, y=f(x)
This also says that
that f is the
constant function.
195
vs
I want you to feel physical pain when you see
∀Inputs x, ∃Java Program J, J(x)=P(x)
196
vs
I want you to feel physical pain when you see
∀Inputs x, ∃Java Program J, J(x)=P(x)
We want to say that
there is a Java program J,
that solves problem P,
on every input I.
Problem P:
• Input: x
• Output: x2
Input:
x
Code:
int J( x )
return(xx)
197
vs
I want you to feel physical pain when you see
∀Inputs x, ∃Java Program J, J(x)=P(x)
But what you are saying is
that each input value can
have its own Java program!
Problem P:
• Input: x
• Output: x2
Input:
1
2
Code: int J1( x )
return(1)
3
int J2( x )
return(4)
int J3( x )
return(9)
4
…
int J4( x )
return(16)
…
Instead
∃Java Program J, ∀Inputs x, J(x)=P(x)
Input:
x
Code:
int J( x )
return(xx)
198
Quantifiers and Games
x, y, y>x
Let’s understand this deeply
For All integer x,
there Exists a y
that is bigger.
No matter how many people are in the hot tub,
x? y is bigger.
you can always fit one more.
0? 1 is bigger.
1? 2 is bigger.
2? 3 is bigger.
3? 4 is bigger.
…
The integers go on forever.
199
Quantifiers and Games
Say, I have a game for you.
We will each choose an integer.
You win if yours is bigger.
I am so nice. I will even let you go first.
Easy.
I choose a trillion trillion.
Well done. That is big!
But I choose a trillion trillion and one
so I win.
Good game.
Let me try again. I will win this time!
200
Quantifiers and Games
You laugh but this is
a very important game
in theoretical computer science.
You choose the size of your Java program.
Then I choose the size of the input.
Likely |I| >> |J|
So you better be sure your Java program
can handle such long inputs.
201
Quantifiers and Games
Assuring
In Predicate logic,
we can state
that I win the game:
all exist
x, y, y>x
I'm an oracle. Let me help.
If I assure you that
all
all boys love, ie, b, Loves(b),
then anyone can give me his favorite boy b
and I assure him that Loves(b) is true.
What about Bob
?
Sure, he is a lover.
202
Quantifiers and Games
Assuring
In Predicate logic,
we can state
that I win the game:
all exist
x, y, y>x
I'm an oracle. Let me help.existIf I assure you that
some boy loves, ie, b, Loves(b),
then I will provide my favorite boy b
and I assure you that Loves(b) is true.
Bob
He is a lover.
203
Quantifiers and Games
Assuring
In Predicate logic,
we can state
that I win the game:
all exist
x, y, y>x
I'm an oracle. Letallmeexist
help. If I assure you that
x, [y, y>x],
then anyone can give me
his favorite integer x
exist
and I assure y, y>x is true.
I choose a x = trillion trillion.
exist
I assure y, y>x.
So I will provide my favorite integer y
and I assure you that y >x is true.
y = trillion trillion and one
Yes y >x
204
Quantifiers and Games
Assuring Proving
In Predicate logic,
we can state
that I win the game:
all exist
x, y, y>x
I'm a prover.
To prove, all I need to do is play the role of the oracle.
For you to prove
that
all
all boys love, ie, b, Loves(b),
then, as the adversary, I give you my favorite boy b
and you must prove Loves(b) is true.
For me to prove
that
exist
some boy loves, ie, b, Loves(b),
then I will construct my favorite boy b
and prove that Loves(b) is true.
205
Quantifiers and Games
Assuring Proving
In Predicate logic,
we can state
that I win the game:
all exist
x, y, y>x
all
exist
I will prove x, [y, y>x].
I choose a x = trillion trillion.
exist
I will prove y, y>x.
I construct y = trillion trillion and one
I will prove y>x.
Surprisingly that
completes the proof.
all
exist
The proof of x, y, y>x.
Let x be an arbitrary integer.
Let y = x+1
Note y = x+1 >x
And this completes the proof.
206
Quantifiers and Games
Assuring Proving
Understanding this game is important!
207
Humans are Mortal
Prove:
∀x, Human(x)Mortal(x)
Human(Socrates)
}
Mortal(Socrates)
“All humans are mortal”
“Socrates is human”
hence “Socrates is mortal”
Aristotle
(384–322 BC)
https://en.wikipedia.org/wiki/Syllogism
208
Humans are Mortal
Prove:
∀x, Human(x)Mortal(x)
Human(Socrates)
}
Mortal(Socrates)
I want to prove (AB)C
As an oracle, I will help.
If AB is false, then (AB)C is automatically true.
Hence, by deduction, we will pretend that AB is true.
During this period of pretending,
I will help by assuring you that AB is true.
This means that each A and B are true.
209
Humans are Mortal
Prove:
∀x, Human(x)Mortal(x)
Human(Socrates)
}
Mortal(Socrates)
I assure you that ∀x, Human(x)Mortal(x)
I assure you that Human(Socrates).
My goal is to prove Mortal(Socrates).
But I am stuck so I will ask my
oracles about who Socrates is.
210
Humans are Mortal
Prove:
∀x, Human(x)Mortal(x)
Human(Socrates)
}
Mortal(Socrates)
I assure you that ∀x, Human(x)Mortal(x)
I am ready for you to give me an x.
I give the oracle x=Socrates.
I assure you that
Human(Socrates)Mortal(Socrates)
I assure you that Human(Socrates).
I use modus ponens to get
Mortal(Socrates).
Excellent. My task is done.
211
Humans are Mortal
Prove:
∀x, Human(x)Mortal(x)
Human(Socrates)
}
Mortal(Socrates)
Formal Proof:
1. Deduction Goal: AB C
2.
AB
3.
A
4.
B
Assumption
Separating And
5.
∀x, Human(x)Mortal(x)
6.
Human(Socrates)Mortal(Socrates)
7.
Human(Socrates)
8.
Mortal(Socrates)
9. AB C
A
Remove ∀
B
Modus Ponens
Deduction Conclude
212
Prove: x, y, x+y=0
Reals
What does this “mean”?
For every x, there is a y such that x+y=0
You can give me more meaning than this!
You get a mark of zero for this!
213
Prove: x, y, x+y=0
Reals
Build understanding by building the statement backwards.
What does each subsequence say about
** value of free variables, * set of values of outer quantifier.
2 -2
x+y=0 “x and y add to zero”
“y is the negative of x”.
½ is the multiplicative inverse of 2.
-2 is the additive
inverse of 2.
“y is the additive inverse of x”.
You can give me more meaning than this!
Maybe some more “formal math”
(that you don’t know)
214
Reals
Prove: x, y, x+y=0 true
Or condensed to x+y(x)=0
y, x, x+y=0 false
Or condensed to x+g =0
y, x, x+y=x true
Or condensed to x+0=x
Build understanding by building the statement backwards.
What does each subsequence say about
** value of free variables, * set of values of outer quantifier.
2 -2
x+y=0 “y is the additive inverse of x”.
y, x+y=0 “x has an additive inverse”.
x, y, x+y=0 “Every real has an additive inverse”.
x+y=0 “x is the additive inverse of y”.
x, x+y=0 “Every real is an additive inverse of y.
y, x, x+y=0 “Some real has every real as an inverse”.
x+y=x “Adding y does not change x”.
x, x+y=x “Adding y makes no changes”.
y, x, x+y=x “There is an additive zero”.
215
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Proof of x, y, x+y=0:
1. Let x be an arbitrary real number.
2. Let y = -x
3. The relation is true.
x+y = x + (-x) = 0
4. Prover can always win.
Hence, the statement is true.
5. “Every real number has an additive inverse.”
216
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Proof of y, x, x+y=0:
The order the players go
REALY matters.
217
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Proof of y, x, x+y=0
1. Let y be ???
2. Let x be arbitrary, eg -y+1
3. The relation is false.
x+y = (-y+1) + y = 1 ≠ 0
4. Adversary can always win.
Hence, the statement is false.
218
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Negations:
[y, x, x+y=0]
y, [x, x+y=0]
y, x, [x+y=0]
y, x, x+y≠0
If there is not a y for which it is true,
then for all y it is not true.
De Morgan's Law
¬(αβ) iff ¬α¬β
Or
=
And
If it is not the case that it is true for all x,
then there is a x for which it is not.
219
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Negations:
[yDogs …]
yDogs
[y≤x …]
y≤x
We are still talking about dogs
and values at most x.
De Morgan's Law
¬(αβ) iff ¬α¬β
Or
=
And
220
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Proof of y, x, x+y=0
1. Let y be ???
2. Let x be arbitrary, eg -y+1
3. The relation is false.
x+y = (-y+1) + y = 1 ≠ 0
4. Adversary can always win.
Hence, the statement is false.
221
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Proof of y, x, x+y≠0:
1. Let y be ???arbitrary
2. Let x be arbitrary, eg -y+1
3. The relation is true.
x+y = (-y+1) + y = 1 ≠ 0
4. Prover can always win.
Hence, the statement is true.
222
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Proof of y, x, x+y=x:
1. Let y be 0
2. x+y = x + 0 = x
No. For Jeff,
you MUST play the game.
223
Reals
Prove: x, y, x+y=0
y, x, x+y=0
y, x, x+y≠0
y, x, x+y=x
true
false
true
true
(additive inverse)
(all additive inverses)
(not all additive inverses)
(additive zero)
Proof of y, x, x+y=x:
1. Let y be 0
2. Let x be arbitrary
3. The relation is true.
x+y = x + 0 = x
4. Prover can always win.
Hence, the statement is true.
5. “There exists an additive zero.”
224
Reals
A: x, y, x+y=0
true (additive inverse)
B:
true (additive zero)
x, x+0=x
C: a,b,c, (a=b) (a+c=b+c)
D:
[ a,b,c, (a+c=b+c) (a=b) ]
Here is a counter example.
[2+infinity=5+infinity] but 2 and 5 are not equal.
•Here is another counter example (with multiplying).
[22=52] mod 6 because 52=10=4 mod 6.
but 2 and 5 are not equal.
225
Reals
A: x, y, x+y=0
true (additive inverse)
B:
true (additive zero)
x, x+0=x
C: a,b,c, (a=b) (a+c=b+c)
D:
[ a,b,c, (a+c=b+c) (a=b) ]
Formal Proof:
1. Deduction Goal: ABC D
2.
ABC
3.
Goal: Proof D
4.
Let a, b, & c be arbitrary
5.
Prove (a+c=b+c) (a=b)
6. ABC D
Assumption
See later
Deduction Conclude
226
Reals
A: x, y, x+y=0
true (additive inverse)
B:
true (additive zero)
x, x+0=x
C: a,b,c, (a=b) (a+c=b+c)
D:
[ a,b,c, (a+c=b+c) (a=b) ]
Formal Proof:
1. Deduction Goal: (a+c=b+c) (a=b)
2.
a+c = b+c
Assumption
3.
c+(-c) = 0
By oracle A
I give you c.
I give you c = -c.
A:
227
Reals
A: x, y, x+y=0
true (additive inverse)
B:
true (additive zero)
x, x+0=x
C: a,b,c, (a=b) (a+c=b+c)
D:
[ a,b,c, (a+c=b+c) (a=b) ]
Formal Proof:
1. Deduction Goal: (a+c=b+c) (a=b)
2.
a+c = b+c
Assumption
3.
c+(-c) = 0
By oracle A
4.
a+c+(-c) = b+c+(-c)
By (2) and oracle C
I give you
a = a+c
b = b+c
c = -c
I let you add –c
onto both sides of
a+c = b+c
C:
228
Reals
A: x, y, x+y=0
true (additive inverse)
B:
true (additive zero)
x, x+0=x
C: a,b,c, (a=b) (a+c=b+c)
D:
[ a,b,c, (a+c=b+c) (a=b) ]
Formal Proof:
1. Deduction Goal: (a+c=b+c) (a=b)
2.
a+c = b+c
Assumption
3.
c+(-c) = 0
By oracle A
4.
a+c+(-c) = b+c+(-c)
By (2) and oracle C
5.
a+0 = b+0
By (3) & (4)
6.
a = b
By oracle B
I give you a.
I give you a+0 = a
I give you b.
I give you b+0 = b
B:
229
Reals
A: x, y, x+y=0
true (additive inverse)
B:
true (additive zero)
x, x+0=x
C: a,b,c, (a=b) (a+c=b+c)
D:
[ a,b,c, (a+c=b+c) (a=b) ]
Formal Proof:
1. Deduction Goal: (a+c=b+c) (a=b)
2.
a+c = b+c
Assumption
3.
c+(-c) = 0
By oracle A
4.
a+c+(-c) = b+c+(-c)
By (2) and oracle C
5.
a+0 = b+0
By (3) & (4)
6.
a = b
By oracle B
7. (a+c=b+c) (a=b)
Deduction Conclude
230
Reals
Hey Adversary,
You can dictate your worst ,
Require it to be as small as you like.
As long as I can counter with an even smaller .
You are on.
The heart of calculous is
231
Reals
Continuous: x,
I claim this function is continuous,
i.e, it does not change values suddenly.
Is it continuous
at value x?
Oh yes.
f
x
232
Reals
Continuous: x,
I’m not convinced.
The value of f seems to be changing.
Well, it is not constant.
Ok. I will allow some change.
Besides, your x' is too far from your x.
I am only talking about local change.
f(x')
f
f(x)
x
x'
233
Reals
Continuous: x, >0, >0, x'[x-,x+], |f(x')-f(x)|≤
I will not allow f to change
by more than my chosen .
Ok. I can honour that.
But then you can’t change
x by more than .
Ok. I will choose
x'[x-,x+],
Then I reassure you that
|f(x')-f(x)|≤
f
f(x)
x'
x
234
Reals
Continuous:
Consider the function f(x) = x2
Prove: x, >0, >0, x'[x-,x+], |f(x')-f(x)|≤
Let x>0 be arbitrary.
Let >0 be arbitrary.
Let = / (2x+ .
Let x'≤x+ be arbitrary.
|f(x')-f(x)| = |(x' 2 - x2| ≤ |(x+ 2 - x2| = |2x+2| = |(2x+)| =
This
choice
looks strange,
Don’t
panic.
butprove,
you will
it isthe
just
what we need.
To
justsee
play
game.
f
f(x)
x'
x
235
Reals
Continuous:
Consider the function below.
Prove: x, >0, >0, x'[x-,x+], |f(x')-f(x)|≤
neg: x, >0, >0, x'[x-,x+], |f(x')-f(x)|>
Let x=5.
Let =0.0001.
Let be arbitrary. It could be .000000000000000000000001
Let x' = x+ = 5+ [x-,x+]
|f(x')-f(x)| = |f(5+ – f(5)| ≤ |3.0002-3| = 0.0002 > 0.0001 =
f(5.1) = 3.0002
f
f(5) = 3
x=5
236
Formal Proof Systems
xP x
P c
for an arbitrary c
xP x
P c for some element c
P c for an arbitrary c
xP x
P c for some element c
xP x
OK, but "arbitrary" vs "some" not well defined.
The line P(c) will mean different things in different contexts.
The book admits that it can cause faulty proofs.
237
237
Review
Don’t panic.
Deeply understand everything
in this review and you will be
fine for the midterm.
238
238
Review
A for And
is OR
→ is Implies
¬ is Not
is parity
T is for True
F is for False
Contrapositive
αβ iff βα iff ¬αβ
Be sure to know the truth tables
for each of these operation.
Given a T/F assignment of the variables
know how to evaluate any expression.
Know how to manipulate expressions
using the rules in the purple table.
De Morgan's Law
¬(αβ) iff ¬α¬β
Double Negation: α iff α
Commutative: αβ iff βα and αβ iff βα
Distributive: γ(αβ) iff (γα)(γβ) and γ(αβ) iff (γα)(γβ)
239
239
Review
A for And
is OR
→ is Implies
¬ is Not
is parity
T is for True
F is for False
Translate between English and Math.
if p, then q
if p, q
p implies q
p is sufficient for q
q if p
q whenever p
q follows from p
q when p
240
p only if q
q is necessary for p
240
Review
A tautology is a sentence that is true
under every setting of the variables.
If you made a table with a row
for each T/F assignment
of the variables p, r, q, ….
the number of rows would be 2# of variables.
We don’t want to check them all.
That is why we have proofs.
241
241
Review
Recall, a proof is a sequence of statements, where each statement is
- either an axiom, i.e. known to be true
- or follows from previous lines using some rule
taught in class, i.e., from the purple table.
Each such rule has the form: From α & β conclude γ.
If you already have lines in your proof of the form α & β.
then you can add the line of the form γ to your proof.
Number the lines of your proof 1, 2, 3, ...
For each, give the name of the rule you use to prove that line.
If you are using previous lines to prove this line,
then give those line numbers.
Do not skip steps. (Except for dropping ¬¬)
Be sure to indent appropriately.
Learn and understand all the rules in the purple table.
242
242
And
Or
And :
Or :
Proof Rules/Lemmas
Techniques/Lemmas
Arguing
Our
Using:
Proving:
From:
Conclude:
Separating And
αβ
α&β
Selecting Or
αβ & α
Cases
αβ, αγ, & βγ
Implies
:
Modus Ponens
α & αβ
Cases
αβ, αγ, & βγ
From:
Conclude:
Eval/Build
α&β
β
αβ
(βγ)
Eval/Build
β
γ
β
γ
Equivalence
α→β & β→α
α iff β
Transitivity
αβ & βγ
αγ
α
α & β
αγ
(αβ)
Excluded Middle
αα & (αα)
Deduction
α→β
Assume α, prove β
Eval/Build
β
α
α & β
γβ
αγ
(αβ)
Contrapositive
αβ iff βα iff ¬αβ
De Morgan's Law
¬(αβ) iff ¬α¬β
By Contradiction: From α(β&β), conclude α.
Double Negation: α iff α
Commutative: αβ iff βα and αβ iff βα Inconsistent: From β and β, conclude anything.
Distributive: γ(αβ) iff (γα)(γβ) and γ(αβ) iff (γα)(γβ)
243
244
245
246
247
248
249
Review
is A for forAll
is E for Exists
Be sure to know what these mean
and the difference between
and
250
250
Quantifiers and Games
Assuring
In Predicate logic,
we can state
that I win the game:
all exist
x, y, y>x
I'm an oracle. Letallmeexist
help. If I assure you that
x, [y, y>x],
then anyone can give me
his favorite integer x
exist
and I assure y, y>x is true.
I choose a x = trillion trillion.
exist
I assure y, y>x.
So I will provide my favorite integer y
and I assure you that y >x is true.
y = trillion trillion and one
Yes y >x
251
Quantifiers and Games
Assuring Proving
In Predicate logic,
we can state
that I win the game:
all exist
x, y, y>x
I'm a prover.
To prove, all I need to do is play the role of the oracle.
252
Quantifiers and Games
Assuring Proving
In Predicate logic,
we can state
that I win the game:
all exist
x, y, y>x
all
exist
I will prove x, [y, y>x].
I choose a x = trillion trillion.
exist
I will prove y, y>x.
I construct y = trillion trillion and one
I will prove y>x.
Surprisingly that
completes the proof.
all
exist
The proof of x, y, y>x.
Let x be an arbitrary integer.
Let y = x+1
Note y = x+1 >x
And this completes the proof.
253
Negations
Negations:
[y, x, x+y=0]
y, [x, x+y=0]
y, x, [x+y=0]
y, x, x+y≠0
If there is not a y for which it is true,
then for all y it is not true.
De Morgan's Law
¬(αβ) iff ¬α¬β
Or
=
And
If it is not the case that it is true for all x,
then there is a x for which it is not.
[yDogs …]
yDogs
[y≤x …]
y≤x
We are still talking about dogs
and values at most x.
254
255
End
(deleted stuff)
256
Arguing Our Rules/Lemmas
Cases
I want to prove that all toys are fun.
I want to play 20-Questions.
I choose a secret toy.
I don’t know which toy truck you have.
But all of these are fun.
No
Alive?
(in the game)
Yes
Machine?
Yes
No
Wheels on road?
Yes
No
Truck?
No
Yes
Fun
257
Arguing Our Rules/Lemmas
Cases
I want to prove that all toys are fun.
I want to play 20-Questions.
I choose a secret toy.
Again!
I don’t know which toy mammal you have.
But all of these are fun.
No
Alive?
(in the game)
Yes
Machine?
Yes
No
Wheels on road?
Yes
No
Truck?
No
Modeled after
real animal?
No
Yes
Mammal?
No Yes
Yes
Fun
Fun
258
Arguing Our Rules/Lemmas
Cases
I want to prove that all toys are fun.
I want to play 20-Questions.
I choose a secret toy.
I know I have proved ALL toys fun
because each and every toy would follow
some path down to a leaf
at which time there is a proof that it is fun.
No
Again!
Alive?
(in the game)
Yes
Machine?
Modeled after
real animal?
No
Yes
Mammal?
Yes
No
Wheels on road?
Yes
No
Truck?
No
Fly?
Yes
No Yes
Fun
Fun
Fun
Fun
Fun
No Yes
Robot?
No
Yes
Fun
Fun
Fun
Fun
259
Arguing Our Rules/Lemmas
Contradiction
We want to prove: n2 is even n is even
To do this we assume “n2 is even”.
But this is hard to use.
Let’s proof the contrapositive: n is odd n2 is odd
(It is not really proof by contradiction, but we need it.)
260
Arguing Our Rules/Lemmas
Contradiction
Prove: n is odd n2 is odd
Assume n is odd
So n=2k+1 for some int k.
So n2 = (2k+1)2 = 4k2 + 4k + 1
= 2(2k2 + 2k) + 1 = 2k' + 1
which is odd.
261
Arguing Our Rules/Lemmas
Proof that √2 is irrational.Contradiction
Let α be “√2 is irrational”.
Let β be “n&d are not both even”
Goal αβ
By way of contradiction assume α, i.e “√2 is rational”.
By definition of rational, √2 = n/d for integers n&d.
If n&d are both even, cancel.
Hence, we can assume “n&d are not both even”
Hence β.
Hence αβ
262
Arguing Our Rules/Lemmas
Proof that √2 is irrational.Contradiction
Let α be “√2 is irrational”.
Let β be “n&d are not both even”
Goal αβ
Again assume α giving √2 = n/d
2d2=n2
n2 is even
n is even
2d2=(2k)2
d2=2k2
d2 is even
d is even
Hence β.
Hence αβ
263
Arguing Our Rules/Lemmas
Proof that √2 is irrational.Contradiction
Let α be “√2 is irrational”.
Let β be “n&d are not both even”
We proved αβ and αβ
Hence α(β&β)
Because α proves a contradiction, α must be wrong.
Hence, α must be true.
Ie “√2 is irrational”.
264
