The Foundations: Logic and Proof, Sets, and
Foundations
PROPOSITIONS
• A proposition is a declarative sentence that is either True or False, but not the
both.
• Example:
•
•
•
•
1. Washington is the capital of United states of America,
2. Toronto is the capital of Canada.
3. 1+1 = 2
4. 2+2 = 3
• Proposition 1 and 3 are True whereas 2 and 4 are False.
The Foundations: Logic and Proof, Sets, and
Foundations
• Consider the following sentences
•
•
•
•
1. What time is it?
2. Read this carefully.
3. x+1 = 2
4. x+y = 3
• 1 and 2 are not Propositions because they are not declarative sentences. 3
and 4 are not Propositions because they are neither True nor False.
• Truth value of proposition is true, denoted by T and truth value of proposition
is false denoted F.
• The area of logic that deals with propositions is called the propositional
calculus or propositional logic.
• Developed by the Greek philosopher Aristotle 2300 years back.
Logical Connectives
Operator
Negation
Conjunction
Disjunction
Exclusive or
Conditional
Biconditional
Symbol Usage






not
and
or
xor
if,then
iff
Compound Propositions: Examples
p = “Cruise ships only go on big rivers.”
q = “Cruise ships go on the Hudson.”
r = “The Hudson is a big river.”
r = “The Hudson is not a big river.”
pq = “Cruise ships only go on big rivers and go on the Hudson.”
pq r = “If cruise ships only go on big rivers and go on the Hudson,
then the Hudson is a big river.”
Negation
This just turns a false proposition to true and the opposite for a true
proposition.
EG: p = “23 = 15 +7”
p happens to be false, so p is true.
Negation – truth table
Logical operators are defined by truth tables –tables which give the
output of the operator in the right-most column.
Here is the truth table for negation:
p
F
T
p
T
F
Conjunction
Conjunction is a binary operator in that it operates on two propositions
when creating compound proposition. On the other hand, negation is
a unary operator (the only non-trivial one possible).
Conjunction
• Conjunction is supposed to encapsulate what
happens when we use the word “and” in English.
I.e., for “p and q ” to be true, it must be the case
that BOTH p is true, as well as q. If one of these is
false, than the compound statement is false as
well.
Conjunction
EG. p = “Clinton was the president.”
q = “Hilary was the president.”
r = “The meaning of is is important.”
Assuming p and r are true, while q false.
Out of pq, pr, qr
only pr is true.
Conjunction – truth table
p
q
p q
T
T
F
F
T
F
T
F
T
F
F
F
Disjunction – truth table
Conversely, disjunction is true when at least one of the components is
true:
p
q
pq
T
T
F
F
T
F
T
F
T
T
T
F
Disjunction – Example
EG. p = “Clinton was the president.”
q = “Hilary was the president.”
r = “The meaning of is is not important.”
Assuming p is true, while q and r are false.
Out of p q, p r, q r
only q r is false.
Review NOT, AND, OR
• What is the disjumction --- or
• What is the conjunction – and
• Negation – not
• Construct the TT for all above
Disjunction – caveat
A: The entrée is served with
soup or salad.
Most restaurants definitely don’t allow you to get both soup and salad
so that the statement is false when both soup and salad is served. To
address this situation, exclusive-or is introduced next.
Exclusive-Or – truth table
p
q
p q
T
T
F
F
T
F
T
F
F
T
T
F
Conditional (Implication)
This one is probably the least intuitive. It’s only
partly akin to the English usage of “if, then” or
“implies”.
DEF: p  q is true if q is true, or if p is false. In the
final case (p is true while q is false) p  q is false.
Semantics: “p implies q ” is true if one can
mathematically derive q from p.
Conditional -- truth table
p
q
p q
T
T
F
F
T
F
T
F
T
F
T
T
Biconditional
Let P and Q be propositions. The biconditional P ↔ Q is the
proposition that is True when P and Q have the same truth values, and
is False otherwise.
P if and only if Q
P→Q and Q→ P
P is necessary and sufficient for Q
Biconditional truth table
P
Q
P↔Q
T
T
F
F
T
F
T
F
T
F
F
T
• Let P and Q be the propositions
P: It is below freezing.
Q: It is snowing.
Express each of these propositions as an English sentences
PQ, PQ, PQ, PQ,
P→Q,
P↔Q
Answers:
It is below freezing and snowing.
It is below freezing but not snowing.
It is not below freezing and it is not snowing.
It is below freezing or snowing.
If it is below freezing then it is snowing.
It is below freezing iff it is snowing
Propositional Equivalences
Tautologies, contradictions, contingencies
DEF: A compound proposition is called a tautology if no
matter what truth values its atomic propositions have, its
own truth value is T.
EG: p  ¬p (Law of excluded middle)
The opposite to a tautology, is a compound proposition
that’s always false –a contradiction.
EG: p  ¬p
On the other hand, a compound proposition whose truth
value isn’t constant is called a contingency.
EG: p  ¬p
Tautologies and contradictions
The easiest way to see if a compound proposition is a
tautology/contradiction is to use a truth table.
p
F
T
p
T
F
p p
p
T
T
F
T
p
T
F
p p
F
F
Tautology example
Demonstrate that
[¬p (p q )]q
is a tautology in two ways:
1. Using a truth table – show that
true
2. Using a proof (will get to this later).
[¬p (p q )]q is always
Tautology by truth table
p q ¬p p q ¬p (p q ) [¬p (p q )]q
T T
T F
F T
F F
Tautology by truth table
p q ¬p p q ¬p (p q ) [¬p (p q )]q
T T
F
T F
F
F T
T
F F
T
Tautology by truth table
p q ¬p p q ¬p (p q ) [¬p (p q )]q
T T
F
T
T F
F
T
F T
T
T
F F
T
F
Tautology by truth table
p q ¬p p q ¬p (p q ) [¬p (p q )]q
T T
F
T
F
T F
F
T
F
F T
T
T
T
F F
T
F
F
Tautology by truth table
p q ¬p p q ¬p (p q ) [¬p (p q )]q
T T
F
T
F
T
T F
F
T
F
T
F T
T
T
T
T
F F
T
F
F
T
Logical Equivalences
DEF: Two compound propositions p, q are logically equivalent if their
biconditional joining p  q is a tautology. Logical equivalence is
denoted by p  q.
EG: The contrapositive of a logical implication is the reversal of the
implication, while negating both components. I.e. the contrapositive
of p q is ¬q ¬p . As we’ll see next: p q  ¬q ¬p
Logical Equivalence of Conditional and
Contrapositive
The easiest way to check for logical equivalence is to
see if the truth tables of both variants have
identical last columns:
p
q
p q
p
q
¬q
¬p
¬q¬p
Q: why does this work given definition of  ?
Logical Equivalence of Conditional and
Contrapositive
The easiest way to check for logical equivalence is to
see if the truth tables of both variants have
identical last columns:
p
q
p q
T
T
F
F
T
F
T
F
T
F
T
T
p
q
¬q
¬p
¬q¬p
Q: why does this work given definition of  ?
Logical Equivalence of Conditional and
Contrapositive
The easiest way to check for logical equivalence is to
see if the truth tables of both variants have
identical last columns:
p
q
p q
p
q
T
T
F
F
T
F
T
F
T
F
T
T
T
T
F
F
T
F
T
F
¬q
¬p
¬q¬p
Q: why does this work given definition of  ?
Logical Equivalence of Conditional and
Contrapositive
The easiest way to check for logical equivalence is to
see if the truth tables of both variants have
identical last columns:
p
q
p q
p
q
¬q
T
T
F
F
T
F
T
F
T
F
T
T
T
T
F
F
T
F
T
F
F
T
F
T
¬p
¬q¬p
Q: why does this work given definition of  ?
Logical Equivalence of Conditional and
Contrapositive
The easiest way to check for logical equivalence is to
see if the truth tables of both variants have
identical last columns:
p
q
p q
p
q
¬q
¬p
T
T
F
F
T
F
T
F
T
F
T
T
T
T
F
F
T
F
T
F
F
T
F
T
F
F
T
T
¬q¬p
Q: why does this work given definition of  ?
Logical Equivalence of Conditional and
Contrapositive
The easiest way to check for logical equivalence is to
see if the truth tables of both variants have
identical last columns:
p
q
p q
p
q
¬q
¬p
¬q¬p
T
T
F
F
T
F
T
F
T
F
T
T
T
T
F
F
T
F
T
F
F
T
F
T
F
F
T
T
T
F
T
T
Q: why does this work given definition of  ?
Logical Equivalences
A: p q by definition means that p  q is a tautology. Furthermore,
the biconditional is true exactly when the truth values of p and of q
are identical. So if the last column of truth tables of p and of q is
identical, the biconditional join of both is a tautology.
Logical Non-Equivalence of Conditional and
Converse
The converse of a logical implication is the reversal of the implication. I.e. the converse of p q
is q p.
EG: The converse of “If Donald is a duck then Donald is a bird.” is “If Donald is a bird then Donald
is a duck.”
As we’ll see next: p q and q p are not logically equivalent.
Logical Non-Equivalence of Conditional and
Converse
p
q
p q q p
(p q)  (q p)
Logical Non-Equivalence of Conditional and
Converse
p
q
T
T
F
F
T
F
T
F
p q q p
(p q)  (q p)
Logical Non-Equivalence of Conditional and
Converse
p
q
T
T
F
F
T
F
T
F
p q q p
T
F
T
T
(p q)  (q p)
Logical Non-Equivalence of Conditional and
Converse
p
q
T
T
F
F
T
F
T
F
p q q p
T
F
T
T
T
T
F
T
(p q)  (q p)
Logical Non-Equivalence of Conditional and
Converse
p
q
T
T
F
F
T
F
T
F
p q q p
T
F
T
T
T
T
F
T
(p q)  (q p)
T
F
F
T
Derivational Proof Techniques
When compound propositions involve more and more atomic
components, the size of the truth table for the compound
propositions increases
Q1: How many rows are required to construct the truth-table of:
( (q(pr ))  ((sr)t) )  (qr )
Q2: How many rows are required to construct the truth-table of a
proposition involving n atomic components?
Derivational Proof Techniques
A1: 32 rows, each additional variable doubles the number of rows
A2: In general, 2n rows
Therefore, as compound propositions grow in complexity, truth tables
become more and more unwieldy. Checking for tautologies/logical
equivalences of complex propositions can become a chore, especially
if the problem is obvious.
Derivational Proof Techniques
EG: consider the compound proposition
(p p )  ((sr)t) )  (qr )
Q: Why is this a tautology?
Derivational Proof Techniques
A: Part of it is a tautology (p p ) and the disjunction of True with any
other compound proposition is still True:
(p p )  ((sr)t ))  (qr )
 T  ((sr)t ))  (qr )
 T
Derivational techniques formalize the intuition of this example.
Tables of Logical Equivalences
• Identity laws
Like adding 0
• Domination laws
Like multiplying by 0
• Idempotent laws
Delete redundancies
• Double negation
“I don’t like you, not”
• Commutativity
Like “x+y = y+x”
• Associativity
Like “(x+y)+z = y+(x+z)”
• Distributivity
Like “(x+y)z = xz+yz”
• De Morgan
Tables of Logical Equivalences
• Excluded middle
• Negating creates opposite
• Definition of implication in
terms of Not and Or
DeMorgan Identities
DeMorgan’s identities allow for simplification of negations of complex
expressions
• Conjunctional negation:
(p1p2…pn)  (p1p2…pn)
• Disjunctional negation:
(p1p2…pn)  (p1p2…pn)
Tautology example (1.2.8.a) Part 2
Demonstrate that
[¬p (p q )]q
is a tautology in two ways:
1. Using a truth table (did above)
2. Using a proof.
Tautology by proof
[¬p (p q )]q
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
Distributive
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
Distributive
ULE
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
 [¬p q ]q
Distributive
ULE
Identity
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
 [¬p q ]q
 ¬ [¬p q ]  q
Distributive
ULE
Identity
ULE
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
 [¬p q ]q
 ¬ [¬p q ]  q
 [¬(¬p) ¬q ]  q
Distributive
ULE
Identity
ULE
DeMorgan
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
 [¬p q ]q
 ¬ [¬p q ]  q
 [¬(¬p) ¬q ]  q
 [p  ¬q ]  q
Distributive
ULE
Identity
ULE
DeMorgan
Double Negation
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
 [¬p q ]q
 ¬ [¬p q ]  q
 [¬(¬p) ¬q ]  q
Distributive
ULE
Identity
ULE
DeMorgan
 [p  ¬q ]  q
Double Negation
 p  [¬q q ]
Associative
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
 [¬p q ]q
 ¬ [¬p q ]  q
 [¬(¬p) ¬q ]  q
Distributive
ULE
Identity
ULE
DeMorgan
 [p  ¬q ]  q
Double Negation
 p  [¬q q ]
Associative
 p  [q ¬q ]
Commutative
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
 [¬p q ]q
 ¬ [¬p q ]  q
 [¬(¬p) ¬q ]  q
Distributive
ULE
Identity
ULE
DeMorgan
 [p  ¬q ]  q
Double Negation
 p  [¬q q ]
Associative
 p  [q ¬q ]
pT
Commutative
ULE
Tautology by proof
[¬p (p q )]q
 [(¬p p)(¬p q)]q
 [ F  (¬p q)]q
 [¬p q ]q
 ¬ [¬p q ]  q
 [¬(¬p) ¬q ]  q
Distributive
ULE
Identity
ULE
DeMorgan
 [p  ¬q ]  q
Double Negation
 p  [¬q q ]
Associative
 p  [q ¬q ]
pT
T
Commutative
ULE
Domination
Exercises
• Construct the truth table for the following Propositions:
•
•
•
•
•
1, (PQ) →Q
2, (PQ) →(PQ)
3, (P→Q) (P→Q)
4, (PQ) (PQ)
5, (P↔Q)(P↔Q)
• Show that the following by constructing the truth table:
• 1, (P↔Q) (P→Q)  (Q→P)
• 2, (P→Q) PQ
• 3, P(Q R)  (PQ) (PR)
Exercises
• Show that (PQ)→(PQ) is a tautology.
• Show that [P(PQ)]→Q is a tautology.
• Determine whether the proposition [P(P → Q)]→Q is a tautology or not.
• Determine whether the proposition [P(P → Q)]→Q is a tautology or not.
• Determine whether the proposition [Q(P → Q)]→ Q is a tautology or
not.
• Show that (P→Q)→Q is a tautology.
• Show that (P→Q)→P is a tautology.
• Show that (P(PQ)) and PQ are logically equivalent.