1. Methods of proof Section 1.6 & 1.7 Wednesday, June 20, 2018
MSU/CSE 260 Fall 2009 1
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2. Proof Methods ¬ p  c T F p c p  c ¬ c  ¬ p p  ¬ c Direct
Wednesday, June 20, 2018
Proof Methods
Direct
Contrapositive
Contradiction p c
p  c
¬ p  c
¬ c  ¬ p
p  ¬ c T F
MSU/CSE 260 Fall 2009 2
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3. How are these questions related?
Wednesday, June 20, 2018
How are these questions related?
Does p logically imply c ?
Is the proposition (p  c) a tautology?
Is the proposition (¬ p  c) is a tautology?
Is the proposition (¬ c  ¬ p) is a tautology?
Is the proposition (p  ¬ c) is a contradiction?
MSU/CSE 260 Fall 2009 3
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4. Proof Methods h1  h2  …  hn  c ?
Wednesday, June 20, 2018
Proof Methods h1  h2  …  hn  c ?
Let p = h1  h2  …  hn . The following propositions are equivalent:
p  c
(p  c) is a tautology. Direct
(¬ p  c) is a tautology. Direct
(¬ c  ¬ p) is a tautology Contrapositive
(p  ¬ c) is a contradiction. Contradiction
MSU/CSE 260 Fall 2009 4
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5. Wednesday, June 20, 2018
Formal Proofs
A proof is equivalent to establishing a logical implication chain
Given premises (hypotheses) h1 , h2 , … , hn and conclusion c, to give a formal proof that the hypotheses imply the conclusion, entails establishing
h1  h2  …  hn  c
We’ll refer to it as the chain method
MSU/CSE 260 Fall 2009
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6. Formal Proof To prove: h1  h2  …  hn  c
Wednesday, June 20, 2018
Formal Proof
To prove: h1  h2  …  hn  c
Produce a series of wffs, p1 , p2 , … pn, c such that each wff pr is:
one of the premises or
a tautology, or
an axiom/law of the domain (e.g., 1+3=4 or x > x+1 )
justified by definition, or
logically equivalent to or implied by one or more propositions pk where 1 ≤ k < r.
p Premise
p Tautology .
pr k, k’, Inf. Rule pn
_____
 c
MSU/CSE 260 Fall 2009 6
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7. “If integer n is odd, then n2 is odd.”
Wednesday, June 20, 2018
Example
Prove the theorem:
“If integer n is odd, then n2 is odd.”
Informal proof:
It is given that n is an odd integer.
Thus n = 2k + 1, for some integer k.
Thus n2 = (2k + 1)2 = 4k2 + 4k = 2(2k2 + 2k) + 1
Therefore, n2 is odd.
MSU/CSE 260 Fall 2009 7
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8. “If integer n is odd, then n2 is odd.”
Wednesday, June 20, 2018
Example h c
Prove the theorem:
“If integer n is odd, then n2 is odd.”
Formal Proof:
1. n is odd Premise (h)
2.  k (n = 2k + 1) Definition of “odd” (Universe is Integers)
3. n = 2c + 1, for some integer c Step 2, specialization
4. (n = 2c + 1)  (n2 = (2c + 1)2) Laws of arithmetic
5. n2 = (2c + 1) Steps 3 & 4, modus ponens
= 4c2 + 4c Laws of arithmetic
= 2(2c2 + 2c) Laws of arithmetic
6.  k (n2 = 2k + 1) Step 5, generalization
7. n2 is odd Definition of “odd”
MSU/CSE 260 Fall 2009 8
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9. Formal Proofs….. Example: Given: p  q, q  r, p.
Wednesday, June 20, 2018
Formal Proofs…..
Example:
Given: p  q, q  r, p.
Prove: r We want to establish the logical implication: (p  q)  (q  r)  p  r.
We can use either of the following approaches
Truth Table
A chain of logical implications
Note that if A  B and B  C then A  C
MSU/CSE 260 Fall 2009 9
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10. Does (p  q)  (q  r)  p  r ? Truth Table Method
Wednesday, June 20, 2018
Does (p  q)  (q  r)  p  r ? Truth Table Method p q r
p  q
q  r T F
MSU/CSE 260 Fall 2009 10
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11. Does (p  q)  (q  r)  p  r ? Chain Method
Wednesday, June 20, 2018
Does (p  q)  (q  r)  p  r ? Chain Method
h1 = p  q, h2 = q  r, h3 = p, c = r We want to prove that h1  h2  h3  c
p Premise
p  q Premise
q Steps 1 & 2, modus ponens*
q  r Premise
r Steps 3&4, modus ponens
*See Table 1, page 66
MSU/CSE 260 Fall 2009 11
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12. Wednesday, June 20, 2018
Example
Prove: If 2 is even and if 3 is even and if the sum of any two even integers is even, then all integers greater than 1 and less than 6 are even.
2 is even Premise
is even Premise
3. n m ((n is even)  (n is even)  (n+m is even)) Premise
4. (2 is even)  (2 is even)  (4 is even) Specialization (twice), step 3, math
5. (3 is even)  (2 is even)  (5 is even) Specialization (twice), step 3, math
is even Conj. & modus ponens, steps 1&4
is even Conj. & modus ponens, steps 1,2,&5
8. (2 is even)  (3 is even)  (4 is even)  (5 is even) Conj.(many times), steps 1,2,6&7
So this is a bona fide theorem – the statement is true!
MSU/CSE 260 Fall 2009 12
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13. Proving conclusions of the form p  q
Direct: Assume p, in addition to the given hypotheses, and conclude q.
Contrapositive: Assume  q, in addition to the given hypotheses, and conclude  p.
Contradiction: Assume both p and  q, in addition to the given hypotheses, and conclude False.
Vacuous: Assume the given hypotheses, and conclude  p.
Trivial: Assume the given hypotheses, and conclude q.
MSU/CSE 260 Fall 2009

14. Example: Direct proof Prove hypothetical syllogism.
Wednesday, June 20, 2018
Example: Direct proof
Prove hypothetical syllogism.
Prove: (p  q)  (q  s)  (p  s)
p Assumption
p  q Premise
q , 2, modus ponens
q  s Premise
s , 4, modus ponens
p  s , 5, direct method of proof
MSU/CSE 260 Fall 2009
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15. Example: Contrapositive proof
Wednesday, June 20, 2018
Example: Contrapositive proof
Prove hypothetical syllogism.
Prove: (p  q)  (q  s)  (p  s)
 s Assumption
q  s Premise
 q , 2, modus tollens
p  q Premise
 p , 4, modus tollens
p  s , 5, contrapositive method
MSU/CSE 260 Fall 2009
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16. Example: Contradiction proof
Wednesday, June 20, 2018
Example: Contradiction proof
Prove hypothetical syllogism.
Prove: (p  q)  (q  s)  (p  s)
p Assumption
 s Assumption
q  s Premise
 q , 3, modus tollens
p  q Premise
 p , 5, modus tollens
p   p , 6, conjunction
False , logical equivalence
p  s , 2, 8, contradiction method
MSU/CSE 260 Fall 2009
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17. Example: Vacuous proof
Prove: If p  r and q   r, then p  q  s
Equivalently, prove: (p  r )  (q   r )  (p  q  s)
p  r Premise
 p  r , Implication
q   r Premise
 q   r , Implication
 p   q , 4, Resolution
 (p  q ) , DeMorgan
p  q  s , Vacuously
MSU/CSE 260 Fall 2009

18. Example: Trivial proof
Prove: If p  r and  r, then q   p
Equivalently, prove: (p  r )  (  r )  ( q   p)
p  r Premise
 r Premise
 p , 2, modus tollens
q   p , Trivially
MSU/CSE 260 Fall 2009

19. Non-examples: Vacuous, trivial proofs
Wednesday, June 20, 2018
Non-examples: Vacuous, trivial proofs
Recall hypothetical syllogism.
Prove: (p  q)  (q  s)  (p  s)
Why didn’t we give example vacuous or trivial proofs of hypothetical syllogism?
MSU/CSE 260 Fall 2009
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20. Wednesday, June 20, 2018
Example
Let h1 = q  d h2 = (q  d )  ¬ p h3 = ¬ p  (a  ¬ b) h4 = (a  ¬ b)  (r  s) c = r  s
we want to establish h1  h2  h3  h4  c.
MSU/CSE 260 Fall 2009 20
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21. Wednesday, June 20, 2018
Solution 1
Let h1 = q  d h2 = (q  d )  ¬ p h3 = ¬ p  (a  ¬ b) h4 = (a  ¬ b)  (r  s) c = r  s we want to establish h1  h2  h3  h4  c.
(q  d )  ¬ p Premise
¬ p  (a  ¬ b) Premise
(q  d )  (a  ¬ b) &2, Hypothetical Syllogism
(a  ¬ b)  (r  s) Premise
(q  d )  (r  s) 3&4, HS
q  d Premise
r  s 5&6, Modus Ponens
MSU/CSE 260 Fall 2009
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22. Wednesday, June 20, 2018
Solution 2
Let h1 = q  d h2 = (q  d )  ¬ p h3 = ¬ p  (a  ¬ b) h4 = (a  ¬ b)  (r  s) c = r  s, we want to establish h1  h2  h3  h4  c.
q  d Premise
(q  d )  ¬ p Premise
¬ p 1&2, and modus ponens
¬ p  (a  ¬ b) Premise
(a  ¬ b) 3&4, modus ponens
(a  ¬ b)  (r  s) Premise
r  s 5&6, modus ponens
MSU/CSE 260 Fall 2009
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23. Example Question: Is [(¬ (p  q))  (¬ p  q)] ≡ (¬ p  q) ?
Wednesday, June 20, 2018
Example
Question: Is [(¬ (p  q))  (¬ p  q)] ≡ (¬ p  q) ?
Different ways to answer the above question
By means of the Truth Table.
By means of derivation.
By formulating it as logical equivalence, that is, as a “proof”.
MSU/CSE 260 Fall 2009 23
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24. Is [(¬ (p  q))  (¬ p  q)] ≡ (¬ p  q) ? Truth Table Method
Wednesday, June 20, 2018
Is [(¬ (p  q))  (¬ p  q)] ≡ (¬ p  q) ? Truth Table Method p q
¬ (p  q)
(¬ p  q)
LHS
RHS
Answer T F
YES
MSU/CSE 260 Fall 2009 24
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25. Is [(¬ (p  q))  (¬ p  q)] ≡ (¬ p  q) ? Derivation Method
Wednesday, June 20, 2018
Is [(¬ (p  q))  (¬ p  q)] ≡ (¬ p  q) ? Derivation Method
(¬ (p  q))  (¬ p  q) ≡ ¬ (¬ (p  q))  (¬ p  q)
≡ (p  q)  (¬ p  q)
≡ ((p  q)  ¬ p)  q )
≡ (¬ p  (p  q))  q ≡ ((¬ p  p)  (¬ p  q))  q
≡ ((T)  (¬ p  q)) q
≡ (¬ p  q) q
≡ (¬ p)  (q q )
≡ (¬ p)  (q)
≡ (¬ p  q)
MSU/CSE 260 Fall 2009 25
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26. Is [ (¬ (p  q))  (¬ p  q)] ≡ (¬ p  q) ? Logical Equivalence Method
Wednesday, June 20, 2018
Is [ (¬ (p  q))  (¬ p  q)] ≡ (¬ p  q) ? Logical Equivalence Method
To show S ≡ R: show that S  R and R  S
In this case, we have S = [¬ (p  q))  (¬ p  q)] and
R = (¬ p  q)
MSU/CSE 260 Fall 2009 26
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27. Prove [(¬ (p  q))  (¬ p  q)]  (¬ p  q)
Wednesday, June 20, 2018
Prove [(¬ (p  q))  (¬ p  q)]  (¬ p  q)
(¬ (p  q))  (¬ p  q) Premise
(p  q)  (¬ p  q) 1, Implication & double neg.
(p  (¬ p  q))  (q  (¬ p  q)) 2, Distribution
(T  q)  (q  ¬ p) , Comm., Assoc, Taut. & Idem.
T  (q  ¬ p) , Domination
¬ p  q , Identity, Commutative
Prove (¬ p  q)  [(¬ (p  q))  (¬ p  q)]
¬ p  q Premise
(¬ (p  q))  (¬ p  q) Trivially, from 1
MSU/CSE 260 Fall 2009 27
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28. Example Is the following reasoning logical?
Wednesday, June 20, 2018
Example
Is the following reasoning logical?
If you are poor then you have no money. If you have money then you are not poor. Therefore, being poor is the same as having no money!
Define the following propositions:
p = “you are poor” q = “you have no money”
Can we conclude that p ≡ q given that p  q and ¬q  ¬p.
In other words, can we prove that: [(p  q)  (¬ q  ¬ p) ]  (p  q) .
MSU/CSE 260 Fall 2009
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29. Solution: Does [(p  q)  (¬ q  ¬ p) ] ≡ (p  q) ?
Wednesday, June 20, 2018
Solution: Does [(p  q)  (¬ q  ¬ p) ] ≡ (p  q) ? p q
p  q
¬ q
¬ p
¬ q  ¬ p
LSH
p  q T F
There is a possibility of not being poor while having no money!
MSU/CSE 260 Fall 2009
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30. Wednesday, June 20, 2018
More Example
Let h1 = p  (q  s) h2 = ¬ r  p h3 = q c = r  s we want to establish h1  h2  h3  c
MSU/CSE 260 Fall 2009 30
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31. Does ( p  (q  s) )  (¬ r  p)  q  (r  s) ?
Wednesday, June 20, 2018
Does ( p  (q  s) )  (¬ r  p)  q  (r  s) ?
¬ r  p Premise
r Assumption
¬ (¬ r ) Double negation, 2
p Disj. syllogism, 1&3
p  (q  s) Premise
q  s Modus ponens, 4&5
q Premise
s Modus ponens, 6&7
r  s Direct proof, 2&8
MSU/CSE 260 Fall 2009 31
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32. General Proof by Contradiction
Proof by contradiction is a general proof method (conclusion can be of any form)
Method:
To prove that h1  h2  …  hn  c: Assume  c, in addition to the hypotheses, and conclude False.
When c has the form p  q, we get the “specialized” version presented earlier.
MSU/CSE 260 Fall 2009

33. Example: General Proof by Contradiction
Wednesday, June 20, 2018
Example: General Proof by Contradiction
Let h1 = q  d h2 = (q  d )  ¬ p h3 = ¬ p  (a  ¬ b) h4 = (a  ¬ b)  (r  s) c = r  s,
Prove by contradiction that
h1  h2  h3  h4  c.
MSU/CSE 260 Fall 2009 33
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34. ¬ (r  s) Contrary Assumption F
Wednesday, June 20, 2018
h1  h  h  h  ¬ c  F (q  d)  ( (q  d )  ¬ p )  (¬ p  (a  ¬ b))  ((a  ¬ b)  (r  s))  ¬ ( r  s)  F
q  d Premise
(q  d )  ¬ p Premise
¬ p 1&2, and modus ponens
¬ p  (a  ¬ b) Premise
(a  ¬ b) 3&4, modus ponens
(a  ¬ b)  (r  s) Premise
r  s 5&6, modus ponens
¬ (r  s) Contrary Assumption F
MSU/CSE 260 Fall 2009 34
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35. Rules of Inference for Predicates
Wednesday, June 20, 2018
Rules of Inference for Predicates
All the Propositional logic rules.
The Universal Specification (US) rule: x P(x)  P(y) for any y in the domain. The rule is also know as Instantiation rule
The Existential Specification (ES) x P(x)  P(y) for some y in the domain.
The Existential Generalization (EG) P(y)  x P(x)
MSU/CSE 260 Fall 2009 35
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36. x (A(x) → B(x)) ≡ x A(x) → x B(x)
Wednesday, June 20, 2018
Other Facts
x (A(x) → B(x)) ≡ x A(x) → x B(x)
x A(x) → x B(x) ≡ x (A(x) → B(x))
x (A(x)  B(x)) ≡ x A(x)  x B(x)
x (A(x)  B(x)) ≡ x A(x)  x B(x)
MSU/CSE 260 Fall 2009 36
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37. Prove that x (H(x) → M(x))  H(s) ⇒ M(s)
Wednesday, June 20, 2018
Prove that x (H(x) → M(x))  H(s) ⇒ M(s)
This is the famous Socrates’s argument All men are mortal Socrates is a man Therefore, Socrates is a mortal
Let H(x) be “x is a man”,
Let M(x) be “x is a mortal” and
Let s be “Socrates”.
MSU/CSE 260 Fall 2009 37
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38. Prove that x (H(x) → M(x))  H(s) ⇒ M(s)
Wednesday, June 20, 2018
Prove that x (H(x) → M(x))  H(s) ⇒ M(s)
x (H(x) → M(x)) Premise
H(s) → M(s) , Universal Specification
H(s) Premise
M(s) &3 and MP
You get to choose s in step 2.
MSU/CSE 260 Fall 2009 38
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39. Prove that x (H(x) → M(x))  x H(x)  x M(x)
Wednesday, June 20, 2018
Prove that x (H(x) → M(x))  x H(x)  x M(x)
x H(x) Premise
H(y) Existential Specification, for some y
x (H(x) → M(x)) Premise
H(y) → M(y) & Universal Specification
M(y) &4, Modus Ponens
x M(x) , Existential Generalization
MSU/CSE 260 Fall 2009 39
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40. Prove that x (A(x)  B(x))  x A(x)  x B(x)
Wednesday, June 20, 2018
Prove that x (A(x)  B(x))  x A(x)  x B(x)
x (A(x)  B(x)) Premise
A(y)  B(y) 1, ES, y is fixed now.
A(y) , Simplification
B(y) , Simplification
x A(x) 3, EG
x B(x) 4, EG
x A(x)  x B(x) 5&6, Conjunction
Question: Is the converse true?
MSU/CSE 260 Fall 2009
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41. Does x A(x)  x B(x)  x (A(x)  B(x)) ?
Wednesday, June 20, 2018
Does x A(x)  x B(x)  x (A(x)  B(x)) ?
x A(x) Premise
A(y) 1, ES, where y is fixed
x B(x) Premise, ES
B(y) 3, ES, where y is fixed
A(y)  B(y) 2 and 4
x (A(x)  B(x)) 5, EG
This proof is invalid. The “y” in step 2 and 4 cannot be assumed to be the same! A new name must be used to denote a (fixed) y’ such
that B(y’) in step 4.
MSU/CSE 260 Fall 2009 41
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42. Prove that x (A(x)  B(x))  x A(x)  x B(x)
Wednesday, June 20, 2018
Prove that x (A(x)  B(x))  x A(x)  x B(x)
¬ (x A(x)  x B(x)) Contrary Assumption
¬ x A(x)  ¬ x B(x) 1 & De Morgan’s
¬ x A(x) 2
x ¬ A(x) 3 & De Morgan’s
¬ x B(x) 2
x ¬B(x) 5 & De Morgan’s
¬A(y) 4, ES, fixed y
¬B(y) , US, free to choose y as in 7
¬A(y)  ¬B(y) 7 & 8
¬ (A(y)  B(y)) 9, De Morgan’s
x (A(x)  B(x)) Premise
A(y)  B(y) , US, any y, same as in 9
Contradiction 10 & 12
MSU/CSE 260 Fall 2009 42
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43. Proofs, and Proof Methods, Summary & Recap
Wednesday, June 20, 2018
Proofs, and Proof Methods, Summary & Recap
What is a logical argument?
Logical Implication (⇒)
When is a mathematical argument correct?
Need rules of inference
MSU/CSE 260 Fall 2009 43
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44. Proof Methods Direct Contrapositive Contradiction Vacuous Trivial
Wednesday, June 20, 2018
Proof Methods
Direct
Contrapositive
Contradiction
Vacuous
Trivial
To disprove, just need a “counterexample”
MSU/CSE 260 Fall 2009 44
© by A-H. Esfahanian. All Rights Reserved.

45. Applications of Logic Day-to-day conversation
Wednesday, June 20, 2018
Applications of Logic
Day-to-day conversation
The equivalent Algebra (Boolean) for circuit design
Program Correctness
Complexity Theory
MSU/CSE 260 Fall 2009 45
© by A-H. Esfahanian. All Rights Reserved.