2. Methods of Proof

3. Definitions
A theorem is a valid logical assertion which can be proved using
Axioms: statements which are given to be true
Rules of inference: logical rules allowing the deduction of conclusions from premises
A lemma is a ‘pre-theorem’ or a result which is needed to prove a theorem.
A corollary is a ‘post-theorem’ or a result which follows directly from a theorem.

4. Rules of Inference H1 H2 Hn   C H1, H2, … Hn are the hypotheses
We use conjunction: H1 ^ H2 ^ H3…
C is the conclusion.
“” means “therefore” or “it follows that”

5. Some Rules of Inference
Addition
Simplification
Conjunction
Modus Ponens
Mode that affirms

6. Example: Simplification
p:”it is below freezing”
q:”it is raining now”
It is below freezing and raining now.
Therefore, it is below freezing.
Simplification

7. Example: Modus Ponens from Latin: mode that affirms
In other words
If the hypothesis p is true
and the hypothesis (p->q) is true
Then I can conclude q

8. Example: Modus Ponens p : “n is greater than 3”
q: “n2 is greater than 9”
Assuming that p is true, and p q is true, then
if is n greater than 3, it follows that n2 is greater than 9.

9. More Rules of Inference
Modus Tollens
mode that denies
Hypothetical Syllogism
Disjunctive Syllogism

10. Exercise
If It rains today, then we will not have a barbecue today. If we do not have a barbecue today, then we will have a barbecue tomorrow
Therefore, if it rains today, then we will have a barbecue tomorrow.

11. Validity of an Argument
An argument is valid if
whenever all hypotheses are true, the conclusion is also true
To prove that an argument is valid:
Assume the hypotheses are true
Use the rules of inference and logical equivalences to determine that the conclusion is true

12. Recap 1.2: Important Equivalences
Identity
p  T  p
p  F  p
Domination
p  T  T
p  F  F
Idempotent
p  p  p
p  p  p
Double Negation
( p)  p

13. Recap 1.2: Important Equivalences
Commutative
p  q  q  p
p  q  q  p
Associative
(p  q)  r  p  (q  r)
(p  q)  r  p  (q  r)
Distributive
p  (q  r)  (p  q)  (p  r)
p  (q  r)  (p  q)  (p  r)
De Morgan’s
(p  q)  p  q
(p  q)  p  q

14. Recap 1.2: Important Equivalences
Absorption
p  (p  q)  p
p  (p  q)  p
Negation
p  p  T
p  p  F

15. Example Consider the following logical argument:
If horses fly or cows eat artichokes, then the mosquito is the national bird. If the mosquito is the national bird then peanut butter tastes good on hot dogs. But peanut butter tastes terrible on hot dogs. Therefore, cows don’t eat artichokes.
Assign propositional variables to each component in the argument

16. Example Assignments: Represent the argument using the variables
p Horses fly
q Cows eat artichokes
r The mosquito is the national bird
s Peanut butter tastes good on hot dogs
Represent the argument using the variables
(p  q)  r
r  s s
 q Conclusion
Hypotheses

17. Example Assertion Reasons 1. (p  q)  r Hypothesis
2. r  s Hypothesis
3. (p  q)  s Hypothetical syll. on 1. and 2.
4. s Hypothesis
5. (p  q) Modus tollens on 3. and 4.
6. p  q DeMorgan on 5.
7. q  p Commutative on 6.
8. q Simplification on 7.
We got our conclusion of “cows don’t eat artichokes”

18. Example Show that the following argument is valid:
It is not sunny this afternoon and it is colder than yesterday. We will go swimming only if it is sunny. If we do not go swimming, then we will take a canoe trip. If we take a canoe trip, then we will be home by sunset. Therefore, we will be home by sunset.

19. Rules of Inference for Quantified Statements
xP(x), then for any
C, therefore P(c) is true
we select any
element c and P(c) is true
Therefore xP(x)
xP(x)
therefore for at
least one specific c,
P(c) is true
we select a particular
element c and P(c) is true
Therefore, xP(x)

20. Example
Everyone in the discrete math class has taken a CS course. Marla is a student in the discrete class. Therefore, Marla has taken a CS course.
D(x): x is in the discrete math class
C(x): x has taken a CS course
x(D(x)  C(x))
D(Marla)
 C(Marla)
Premises
1. x(D(x)  C(x)) Premise
2. D(Marla)  C(Marla) Univ. Inst. using 1.
3. D(Marla) Premise
4. C(Marla) Modus ponens using 2. and 3.

21. Exercise
A student in this class has not read the book. Everyone in this class passed the first exam.
Therefore, Someone who passed the first exam has not read the book

22. Fallacies
Fallacies resemble rules of inference but are based on contingencies rather than tautologies. They are incorrect inferences.
Three common fallacies
Affirming the Consequent
Denying the Hypothesis
Circular Reasoning (begging the question)

23. Fallacy of Affirming the Consequent
((p  q)  q)  p is not a tautology and therefore not a rule of inference.
If you do every problem in this book, then you will learn discrete mathematics. You learned discrete mathematics. Therefore, you did every problem in this book.
Exercise. Compare with modus ponens and tollens

24. Fallacy of Denying the Hypothesis
((p  q)  p)  q is not a tautology and therefore not a rule of inference.
If you do every problem in this book, then you will learn discrete mathematics. You did not do every problem in this book. Therefore, you did not learn discrete mathematics.
Exercise. Compare with modus ponens and tollens

25. Circular Reasoning
One or more steps of the proof are based upon the truth of the statement being proved.
Also known as begging the question

26. Exercises
1, 3, 4, 7, 9, 11

27. Recap: Some Rules of Inference
Addition
Simplification
Conjunction
Modus Ponens
Mode that affirms

28. Recap: More Rules of Inference
Modus Tollens
mode that denies
Hypothetical Syllogism
Disjunctive Syllogism

29. Rules of Inference for Quantified Statements
xP(x), then for any
C, therefore P(c) is true
we select any
element c and P(c) is true
Therefore xP(x)
xP(x)
therefore for at
least one specific c,
P(c) is true
we select a particular
element c and P(c) is true
Therefore, xP(x)

30. Methods of Proof Direct proof Indirect proof Vacuous proof
Trivial proof
Proof by contradiction
Proof by cases
Existence proof

31. Proof Basics p q p q We want to establish the truth of p  q
p may be a conjunction of other hypotheses
p  q is a conjecture until a proof is produced F T p q
p q

32. Direct Proof Assume the hypotheses are true
Use rules of inference and any logical equivalences to establish the truth of the conclusion
HOW TO PROVE:
If p is true ,then q has to be true for p—>q to be true
Example: The proof we did earlier about cows not eating artichokes was an example of a direct proof

33. Example
Give a direct proof of the theorem: “If n is an odd integer, then n2 is an odd integer”
(n is odd)  (n2 is odd)
Using the following definition:
If n is even, then exist an integer k such that n=2k, and It is odd, if there exist and integer k such that n=2k+1.

34. Example (Cont) Assume the hypothesis “n is odd” true:
Since n is odd, then k n=2k+1
Now, is the conclusion “n2 is odd” true?
n2 = (2k+1)2 = 4k2 +4k +1
= 2(2k2+2k)+1
= 2 (m) +1, where some integer m=2k2+2k
Since n2 = 2(m)+1, then “n2 is odd” is true
Proof complete

35. Indirect Proof A direct proof of the contrapositive Proof ~q  ~p
Remember: pq is equivalent to ~q  ~p
Proof ~q  ~p
Assume that q is true i.e., q is false
Use rules of inference and logical equivalences to show that p is true i.e., p is false

36. Example
Give an indirect proof to the theorem “if 3n+2 is odd, then n is odd”
(3n+2 is odd)  (n is odd)
p: 3n+2 is odd, ~p: 3n+2 is even
q: n is odd, ~q: n is even
The contrapositive is:
~(n is odd)  ~ (3n+2 is odd) , in other words
(n is even)  (3n +2 is even)

37. Example (Cont)
Assuming the hypothesis (of the contra positive) “n is even” true
Then n=2k
Now, is the conclusion (of the contrapositive) “3n+2 is even” true?
3n+2 = 3(2k)+2=6k+2
=2(3k+1)
=2(m), where m =3k+1
Then “3n+2 is even” is true
Proof complete

38. Vacuous Proof
If we know one of the hypotheses in p is false then pq is vacuously true.
F  T and F  F are both true.
Example:
If I am both rich and poor, then hurricane Katrina was a mild breeze.
The hypotheses (pp) form a contradiction, therefore q follows from the hypotheses vacuously.
Sometimes used to proof theorems using UNIVERSAL QUANTIFICATION

39. Example
Show that P(0) is true where P(n): If n > 1, then n2 > n.
P(n): (n>1)  (n2 > n)
P(0): (0>1)  ( 02 > 0)
Since the hypothesis (0>1) is false, P(0) is automatically true.
Note that we do not even pay attention to the conclusion “02 > 0”

40. Trivial Proof If we know q is true, then pq is true
F  T and T  T are both true.
Example:
If it’s raining today then the empty set is a subset of every set.
The assertion is trivially true independent of the truth value of p.

41. Example
Show that P(0) is true where P(n): If a  b > 0, then an  bn.
P(n): (a  b > 0)  (an  bn)
P(0): (a b > 0)  (a0  a0), in other words
P(0): (a b > 0)  (1  1),
Since the conclusion (1  1) is true, hence P(0) is true.
Note that we do not even pay attention to the hypothesis “(a b > 0) “

42. Proof by Contradiction (reduction to the absurd )
We want to proof that pq, but… what if we can proof that ~p implies a contradiction (i.e., q is FALSE no matters what, or an absurd)??
Mathematical definition of the proof
Find a contradiction q such that
pq  pF  T
Consequently, if we show the contradiction, then the assumption ~p is wrong (FALSE), so p must be true
To prove that p is true, you have to show that p leads to a contradiction i.e., you have to prove that pF is true.

43. Example Prove that √2 is irrational
P: “√2 is irrational”
What if ~p is true, “√2 is rational”
Does this lead to a contradiction???
If √2 is rational, then a,b integers such that √2 =a /b (assuming r:”a and b have no common factors”)
√2 = a/ b , then 2 = a2/b2, then 2b2=a2

44. Example cont Therefore, ~p is false, p is true
Since a2=2 (b2), then a2 is even, therefore a is even, then
2b2 = (2c)2, for some integer c
2b2=4c2, so b2=2c2
Thus, b is even too.
If a and b are even, then they have at least one common factor (2), so the assumption r is contradicted: p-> (r^~r)
Therefore, ~p is false, p is true
“√2 is irrational is true

45. Proof by Contradiction (Cont..)
An indirect proof of an implication pq can be rewritten as a proof by contradiction.
Assume that both p and q are true.
Then use a direct proof to show that
q  p
This leads to the contradiction pp.
HOMEWORK. Example 22 (pg 67) If 3n+2 is odd, then n is odd.

46. Proof by Cases
Break the premise of pq into an equivalent disjunction of the form p1p2pn
Then use the equivalence
[(p1p2pn)  q]  [(p1q)(p2 q)   (pnq)]
Each of the implications pi q is a case.
You must
Convince the reader that the cases are inclusive (i.e., they exhaust all possibilities)
Establish all implications

47. Proof by Cases (Cont..) If n is an integer, then n2+1 is positive.
p: n is an integer
q: n2+1 is positive
p  (p1p2p3)
where, p1: n = 0
p2: n > 0
p3: n < 0
Now prove that ((p1q)(p2 q)(p3 q)) is true i.e., all the cases (p1q), (p2 q), and (p3 q) are true.

48. Proving an equivalence
To prove that pq, you need to show that pq is a tautology.
You can do that by showing that pq and qp are both true since,
pq  [(pq)(qp)]
Example: The integer n is odd if and only if n2 is odd.

49. Existence Proof The proof of xP(x) is called an existence proof.
Constructive
Non-constructive
Constructive existence proof
Find an element c in the universe of discourse such that P(c) is true
Non-constructive existence proof
Do not find c, rather, somehow prove xP(x) is true
Generally, by contradiction
Assume no c exists that makes P(c) true
Derive a contradiction