Proof by Deduction
Deductions and Formal Proofs
• A deduction is a sequence of logic statements, each
of which is known or assumed to be true
• A formal proof of a conclusion C is a deduction that
ends with C
• What is known or assumed to be true comes in three
variations:
– Premises that are assumed to be true
– Any statement known to be true (e.g., 1+1 = 2)
– Sound rules of inference that introduce logic statements
that are true (assuming the statements they are based on are
also true)
Careful Again…
• A formal proof guarantees that the original
statement is a tautology
– If any premise is false, the statement is true (P  Q is
true when P is false)
– If all premises are true, we are able to guarantee that the
conclusion is true (by the proof derivation)
– Thus, the statement is always true (i.e., is a tautology)
• HOWEVER:
– A proof does NOT guarantee that the conclusion is
true!
Valid Proof vs. Valid Conclusion
Consider the following proof:
If 2 > 3, 2 > 3  3 < 2, then 3 < 2 .
1. 2 > 3
2. 2 > 3  3 < 2
3. 3 < 2
premise
premise
1&2, modus ponens
Yes! Sound argument
Is the proof valid?
Is the conclusion valid? No!
Hence, a valid proof does not guarantee a valid
conclusion. What a valid proof does guarantee is that
IF the premises are all true, then the conclusion is too.
Main Rules of Inference
A, B |= A  B
Combination
A  B |= B
Simplification
A  B |= A
Variant of simplification
A |= A  B
Addition
B |= A  B
Variant of addition
A, AB |= B
Modus ponens
B, AB |= A
Modus tollens
AB, BC |= AC
Hypothetical syllogism
A  B, A |= B
Disjunctive syllogism
A  B, B |= A
Variant of disjunctive syllogism
AB, AB |= B
Cases
AB |= AB
Equivalence elimination
AB |= BA
Variant of equivalence elimination
AB, BA |= AB
Equivalence introduction
A, A |= B
Inconsistency law
Reduction of Rules of Inference
Easy to show validity of rules with truth tables
• Using laws, we can also reduce the number of
rules of inference
– Too many to remember
– Too many to program! (Project)
• Example: modus tollens
B
AB
A
B  A contrapositive
B
A
modus ponens
It turns out that modus ponens plus some simple laws are usually
enough, and it can get even simpler…
1. Exhaustive Truth Proof
Show true for all cases
– e.g.
Prove x2 + 5 < 20 for integers 0  x  3
02 + 5 = 5
12 + 5 = 6
22 + 5 = 9
32 + 5 = 14
< 20
< 20
< 20
< 20
T
T
T
T
– Thus, all cases are exhausted and true
– Same as using truth tables  when all
combinations yield a tautology
2. Equivalence to Truth
2a.
Transform logical expression to T
Prove: R  S  (R  S)
R  S  (R  S)
 R  S  R  S
RSRS
 (R  S)  (R  S)
T
de Morgan’s law
double negation
implication (PQ  PQ)
law of excl. middle (P  P)  T
Use laws only!
2. Equivalence to Truth (cont’d)
2b. Transform lhs to rhs or vice versa
Prove: P  Q  P  Q  Q
P  Q  P  Q
 (P  P)  Q
TQ
Q
distributive law (factoring)
law of excluded middle
identity
Use laws only!
3. If and only if Proof
• Also called necessary and sufficient
• Since P  Q  (P  Q)  (Q  P), we can
create two standard deductive proofs:
–PQ
–QP
• Transforms proof to a standard deductive
proof  actually two of them
3. If and only if Proof: Example
Prove: 2x – 4 > 0 iff x > 2.
Thus, we can do two proofs:
(1) If 2x – 4 > 0 then x > 2.
(2) If x > 2 then 2x – 4 > 0.
Proof:
(1) 1. 2x – 4 > 0 premise
2. 2x > 4
1, algebraic equiv.
3. x > 2
2, algebraic equiv.
(2) Similarly, suppose x > 2. Then 2x > 4 and thus 2x – 4 > 0.
Note that we could have also converted the lhs to the rhs
Proof: 2x – 4 > 0  2x > 4  x > 2
4. Contrapositive Proof
• Since P  Q  Q  P (by the law of
contrapositive), we can prove either side of
the equivalence, whichever is EASIER
• E.g., prove: if s is not a multiple of 3, then s
is not a solution of x2 + 3x – 18 = 0
– Infinitely many non-multiples of 3 and nonsolutions to the quadratic equation – HARD!
– Let:
• P = s is a multiple of 3
• Q = s is a solution of x2 + 3x – 18 = 0
– Instead of P  Q, we can prove Q  P
4. Contrapositive Proof (cont’d)
Prove: if s is a solution of x2 + 3x – 18 = 0,
then s is a multiple of 3
Proof: The solutions of x2 + 3x – 18 = 0 are 3
and – 6. Both 3 and – 6 are multiples of 3.
Thus, every solution s is a multiple of 3.
Note: Using a contrapositive proof helps get rid of the not’s
and makes the problem easier to understand  indeed, the
contrapositive turns this into a simple proof we can do
exhaustively.
5. Proof by Contradiction
• Note:
P
P 
P  F
T
T
F
T
F
T
T
F
• Thus, P  P  F
• Substituting R  S for P, we have
• RS

(R  S)  F
• And finally
(R  S)  F  (R  S)  F
 R  S  F
 R  S  F
implication
de Morgan’s
double negation
5. Proof by Contradiction (cont’d)
• Thus, we have
– RS
 R  S  F
• Hence, to prove R  S, we can
– Negate the conclusion
– Add it as a premise (to R)
– Derive a contradiction (i.e. derive F)
• Any contradiction (e.g., P  P) is equivalent to F, so
deriving any contradiction is enough
Note: in our project we will implement the semantics of our language by
programming the computer to do a proof by contradiction
5. Proof by Contradiction: Example
• Prove: the empty set is unique
Since P  (P  F), we can prove P  F instead
i.e. We can prove: if the empty set is not unique, then there is a
contradiction, or, in other words, assuming the empty set is not
unique leads (deductively) to a contradiction.
• Proof:
– Assume the empty set is not unique
– Then, there are at least two empty sets 1 and 2 such that
1  2
– Since an empty set is a set and an empty set is a subset of
every set, 1  2 and 2  1 and thus 1 = 2
– But now we have 1  2 and 1 = 2  a contradiction!
– Hence, the empty set is unique