1. Discrete Maths 4. Proofs Objectives
, Semester 2,
4. Proofs
Objectives
to introduce proofs: assumptions, conclusions, rules of inference; different ways of proving if-then statements

2. Overview 1. What is a Proof? 9. Proving a Biconditional
2. Rules of Inference
3. Using Rules to Build a Proof
4. More Inference Rules
5. More Examples
6. More Proof Jargon
7. Two Simple Ways to Prove (p → q)
8. More Advanced Proofs of (p  q)
9. Proving a Biconditional
10. Proof by Cases
11. Existence Proofs
12. Counter-examples
13. More Information

3. 1. What is a Proof? Two assumptions /premises / statements:
“All men are mortal.”
and
“Socrates is a man.”
Prove the conclusion:
“Socrates is mortal.”
the steps in going from
assumptions to a conclusion
is called a proof
proof
assumptions
conclusion

4. In Logic Rules of inference are used to create a proof assumptions
Pred. Sec 4.1
assumptions
how do
we prove
this?
conclusion
Rules of inference are used to create a proof
most rules work with both propositional and predicate logic

5. 2. Rules of Inference For proofs using propositional logic:
Modus Ponens
Modus Tollens
Transitivity
Disjunctive Syllogism
Addition
Simplification
Conjunction
Resolution
rules involving 
(used a lot)
used in the Prolog
logic programming
language

6. 2.1. Modus Ponens Example Let p be “It is snowing.”
assumptions
conclusion
Example
Let p be “It is snowing.”
Let q be “I study discrete math.”
p  q: “If it is snowing, then I study discrete math.”
p: “It is snowing.”
q: “I study discrete math.”

7. Why does Modus Ponens work?
Look at the truth table for 
assumptions T ?
p  q q T p
There is only
one possible value
for q, true.

8. 2.2. Modus Tollens Example Let p be “it is snowing.”
Let q be “I study discrete math.”
p  q: “If it is snowing, then I study discrete math.”
q: “I do not study discrete math.”
p: “It is not snowing.”

9. Why does Modus Tollens work?
Look at the truth table for 
assumptions T ?
p  q p T q
There is only
one possible value
for p, false.
This means p == T
(i.e. q == F)

10. 2.3. Transitivity (chaining)
This may be why math
people use the  symbol
2.3. Transitivity (chaining)
assumptions p q r
Example
Let p be “it snows.”
Let q be “I study discrete math.”
Let r be “I will get an A.”
p  q: “If it snows, then I study discrete math.”
q  r: “If I study discrete math, then I will get an A.”
p  r: “If it snows, then I will get an A.”
conclusion

11. T
p  q
This shows why using truth tables for proofs does not scale up. T
q  r
possible
values
for p, r R ?
p  r

12. 2.4. Disjunctive Syllogism
p  q T p ??
the “get part of “ rule q
Example
Let p be “I study discrete math.”
Let q be “I study English literature.”
p  q: “I study discrete math or I study English literature.”
p: “I do not study discrete math.”
q: “I study English literature.”

13. 2.5. Addition Example Let p be “I study discrete math.”
the “add anything “ rule ??
p  q
Example
Let p be “I study discrete math.”
Let q be “I will visit the Moon.”
p: “I study discrete math.”
p  q: “I study discrete math or I will visit the Moon.”

14. 2.6. Simplification p Example Let p be “I study discrete math.”?
2.6. Simplification T q
p  q p ?
Two rules p
the “get part of “ rule
Example
Let p be “I study discrete math.”
Let q be “I study English literature.”
p  q: “I study discrete math and English literature”
q: “I study discrete math.”

15. 2.7. Conjunction Example Let p be “I study discrete math.”q ?
p  q
the “combine two
statements with “ rule -- useful
Example
Let p be “I study discrete math.”
Let q be “I study English literature.”
p: “I study discrete math.”
q: “I study English literature.”
p  q: “I study discrete math and I study English literature.”

16. 2.8. Resolution Graphically: p  r p  q q  r Example
Let p be “I study discrete math.”
Let r be “I study English.”
Let q be “I study databases.”
p  r: “I do not study discrete math or I study English.”
p  q: “I study discrete math or I study databases.”
q  r: “I study databases or I study English.”

17. 3. Using Rules to Build a Proof
Each statement (line) is either an assumption or a new statement made by using a rule of inference. S1 S2 . Sn C
assumptions or
new statements
and
this sequence of statements is
the proof
conclusion

18. Example 1 Simplification Simplification Assumption:
Show that q is a conclusion.
Simplification
Simplification

19. Example 2 continued Four Assumptions:
“It is not sunny and it is colder than yesterday.”
“We will go swimming 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 night.”
Show the conclusion:
“We will be home by night.”
a q if p
example
continued

20. Create propositional variables:
p: “It is sunny.” r: “We will go swimming.”
t: “We will be home by night.”
q: “It is colder than yesterday.” s: “We will take a canoe trip.”
Translation problem into propositional logic:
Four assumptions:
p  q r  p r  s s  t
Conclusion: t

21. 3. Construct the proof
Note that a truth table would have 32 rows since we have 5 propositional variables.

22. 4. More Inference Rules Rules using variables and quantification:
for predicate logic proofs
Rules using variables and quantification:
Universal Instantiation (UI)
Universal Generalization (UG)
Existential Instantiation (EI)
Existential Generalization (EG)
Universal Modus Ponens (MP)
Universal Modus Tollens (MT)
involving 
involving 
versions of same-named
propositional rules

23. 4.1. Universal Instantiation (UI)
the “choose any element“ rule U P a b
... c
Example
The domain U is all dogs.
x P(x): “All dogs are cute.”
P(c): “Dog c is cute.”
all elements are in P;
nothing outside

24. Universial instantiation can be used on any expression surrounded by .
For example:
x P(x)  Q(x) y P(y)  Q(y)
 P(c)  Q(c)  P(a)  Q(a)

25. 4.2. Universal Generalization (UG)P a
any c b
... c
all elements are in P;
nothing outside
If you always choose a dog that is cute.
“Therefore, all dogs are cute.”

26. 4.3. Existential Instantiation (EI)U P a b
... c d
...
some elements are in P;
others are outside
the “choose an element carefully“ rule

27. Existential instantiation can be used on any expression surrounded by .
But you must be careful to select an element that is covered by the logic (i.e. inside the set).
For example:
x P(x)  Q(x) y P(y)  Q(y)
 P(c)  Q(c)  P(a)  Q(a)

28. 4.4. Existential Generalization (EG)U P a b
... c d
...
some elements are in P;
others are outside
Example:
P(c): “Student c got an A in the class.”
x P(x): “Someone got an A in the class.”

29. 4.5. Universal Modus Ponens (MP)
x (P(x)  Q(x)
P(a)
 Q(a) P Q a
‘a’ must be in here
according to the assumptions
See the Socrates example , in the next few slides.

30. 4.6. Universal Modus Tollens (MT)
x (P(x)  Q(x)
Q(a)
 P(a) P Q a
‘a’ must be in here
according to the assumptions

31. 5. More Examples Example 1: Prove the conclusion Assumptions:
“John Smith has two legs”
Assumptions:
“Every man has two legs.” “John Smith is a man.”
Convert to predicate logic:
M(x) is “x is a man”
L(x) is “ x has two legs”
John Smith (J) is in the domain.
M(J)
L(J)
x M(x)  L(x)
Pred. Sec 4.1
continued

32. U M L J
Proof:

33. Example 2 Prove the conclusion Assumptions:
“Someone who passed the first exam has not read the book.”
Assumptions:
“Some students in this class have not read the book.”
“Everyone in this class passed the first exam.”
continued

34. Convert to predicate logic: C(x) is “x is in this class”
B(x) is “x has read the book”
P(x) is “x passed the first exam.”
Translate assumptions and conclusion:
Pred. Sec 4.3
Pred. Sec 4.1
continued

35. Proof

36. Socrates Example "All men are mortal" "Socrates is a man"
Pred. Sec 4.1
"All men are mortal"
"Socrates is a man"
"Socrates is mortal"
Proof

37. 6. More Proof Jargon
A theorem is a conclusion shown to be true by writing a proof.
A lemma is a statement that helps to prove a theorem.
A corollary of a theorem is another statement that you can prove using the theorem.
A conjecture is a conclusion that might be true. Once you have proved it, then it becomes a theorem.

38. 7. Two Simple Ways to Prove (p → q)
1. Trivial Proof: if q is true, then p → q is true.
e.g. “If it is raining then 1=1.”
this statement
(p  q) is true p q
look at the
cases in the
truth table
p doesn't affect
the truth of
(p  q)

39. 2. Vacuous Proof: if p is false then p → q is true
e.g. “If I am both rich and poor, then = 5.”
this statement
(p  q) is true q p
q doesn't affect
the truth of
(p  q)
look at the
cases in the
truth table

40. 8. More Advanced Proofs of (p  q)
There are three main advanced techniques:
direct proof
proof by contrapositive (contraposition)
proof by contradiction

41. 8.1. Direct Proof of p → q Assume that p is true.
Use rules of inference to show that q is true.
If p and q are true then (p q) is true
look at the cases in the
truth table

42. Example DP-1 Prove “If n is an odd integer, then n2 is odd.”
Assume that n is odd (i.e. p is true).
Then n = 2k + 1 for some integer k.
Squaring both sides:
n2 = (2k + 1)2 = 4k2 + 4k +1 = 2(2k2 + 2k) + 1= 2r + 1
So n2 is an odd integer (i.e. q is true).
This means that pq is true. p q

43. Example DP-2
Definition: A real number x is rational if there exists integers a and b where b≠0 such that x = a/b, and a and b have no common factors.
Prove that the sum of two rational numbers is rational.
As an if-then:
if r is rational and s is rational then (r+s) is rational.
Assume that r and s are rational (i.e. p is true). p q
continued

44. This shows that the sum (r +s) is rational (i.e. that q is true)
Since r and s are rational, we can use the definition on the previous slide:
This shows that the sum (r +s) is rational (i.e. that q is true)
This means that pq is true.
where v = pu + qt
w = qu ≠ 0

45. 8.2. Prove by Contrapositive
Instead of proving (p  q), we prove its contrapositive, which is (q  p)
(p  q) and (q  p) are logically the same (look at their truth tables), but sometimes it is easier to build a proof for (q  p).

46. Proof Steps for (q → p)
Assume that q is true.
Use rules of inference to show that p is true.
If q and p are true (i.e. that q and p are false) then (q → p) is true
see the truth table on the previous slide

47. Example CP-1
Prove that if n is an integer and 3n + 2 is odd, then n is odd.
Prove the contrapositive (q → p) is true
This is: if (n is not odd) then (3n + 2 is not odd)
Simplify using the mathematical meaning of odd/even: if (n is even) then (3n + 2 is even) p q q p q p
continued

48. Assume n is even (i.e. that q is true). So, n = 2k for some integer k. Thus 3n + 2 = 3(2k) + 2 =6k +2 = 2(3k + 1) = 2j
Therefore 3n + 2 is even (i.e. p is true).
We have shown (¬q → ¬p ) is true, so (p → q) is true.

49. Example CP-2 Prove that if n2 is odd, then n is odd. continued
Prove the contrapositive (q → p) is true
This is: if (n is not odd) then (n2 is not odd)
Simplify using the mathematical meaning of odd/even: if (n is even) then (n2 is even) p q q p q p
continued

50. Assume n is even (i. e. q is true)
Assume n is even (i.e. q is true). Therefore, there is an integer k such that n = 2k. Square both sides to get n2 = 4k2 = 2 (2k2)
This means that n2 is even (i.e. p is true)
We have shown (¬q → ¬p ) is true, so (p → q) is true.

51. 8.3. Prove by Contradiction
Instead of proving that (p  q) is true, prove the negative is false:
i.e. prove that (p  q) is false
Transform: prove (p  q) is false  prove (p  q) is false
In words this means that we must show that p and q cannot both be true.

52. Example CD-1
Prove that if you pick 22 days from the calendar, at least 4 must fall on the same day of the week.
p  q: if (pick 22 days) then (≥ 4 are same day)
Show that (p  q) is false i.e. show that (p  q) is false
q = (≥ 4 are same day) = (≤ 3 are same day) p q
continued

53. p = pick 22 days = 3 weeks + 1 day
If the pick includes 3 weeks then each day is repeated 3 times.
"+ 1 day" means that one of the days occurs 4 times.
This is not the same as q
In other words, (p  q) is false

54. Example CD-2 Prove that √2 is irrational.
This famous problem is not of the form (pq), but we can use a slight variation of proof by contradiction.
We want to prove that p is true. Instead we will prove that the negative is false:
i.e. prove that p is false
Assume p: √2 is not irrational  √2 is rational p
continued

55. If √2 is rational then √2 = a/b, where b≠ 0 and a and b have no common factors (see slide 43).
Therefore a2 is even.
If a2 is even then a is even (we should really prove this).
Since a is even, a = 2c for some integer c.
Therefore b2 is even. So b is even.
continued

56. Since a and b are both even, then a and b have a common factor of 2
This contradicts p which includes the assumption that a and b have no common factors.
So p is false, which means that p is true.
Therefore, √2 is irrational.

57. 9. Proving a Biconditional (pq)
Prove (p ↔ q) is true in two steps (cases):
show that (p → q ) is true and
show that (q →p) is true
Why? Because (p ↔ q)  (p → q)  (q →p)
As shown in the
truth table p q
pq
qp
pq 
p ↔q T F

58. For each of the two cases, you can use any proof technique you like:
e.g. trivial, vacuous, direct, contrapositive, contradiction

59. Example Prove “n is odd if and only if n2 is odd.” Two cases: 
p  q: if (n is odd) then (n2 is odd)
q  p: if (n2 is odd) then (n is odd)
we proved this case in example CP-2 p  q

60. 10. Proof by Cases We want to prove the complex implication:
We can rewrite this as a conjunction (and) of simple implications:
((p1  q)  (p2  q)   (pn  q))
Each of the implications is a case that needs to be proved:
show (p1  q) is true, show (p2  q) is true, etc.

61. How to rewrite: Remember that (p  q)  (p  q)
(p1  p2  ...  pn) q  (p1  p2  ...  pn)  q  (p1  p2  ...  pn)  q // De Morgan's  (p1  q)  (p2  q)  ...  (pn  q) // distribution  (p1  q)  (p2  q)  ...  (pn  q)

62. Example
Define max(a, b) = a if a ≥ b, otherwise = b if (a, b, c are real numbers) then max( max(a,b), c) == max(a, max(b, c)) Where are the cases inside p (i.e. p1, p2, ..pn)? They are all the different possible values for a, b, and c.
p (complex) q
continued

63. There are only 6 p cases that we need to look at: a ≥ b ≥ c a ≥ c ≥ b
b ≥ a ≥c
b ≥ c ≥a
c ≥ a ≥ b
c ≥ b ≥ a
cases p1, p2, ..., p6

64. In other words, we must prove:
(p1  q) if (a ≥ b ≥ c) then max( max(a,b), c) == max(a, max(b, c))
(p2  q) if (a ≥ c ≥ b) then max( max(a,b), c) == max(a, max(b, c)) :
(p6  q) if (c ≥ b ≥ a) then max( max(a,b), c) == max(a, max(b, c))
and
and

65. Case 1 Proof
I'll just look at case 1 (p1  q), since the other cases are similar.
Use direct proof:
Assume that p1 is true.
Use rules of inference to show that q is true.
If p1 and q are true then (p1 q) is true

66. Assume a ≥ b ≥ c (i.e. p1 is true)
Plug this information into max():
max(a, b) = a max(a, c) = a max(b, c) = b
This means that
max(max(a, b), c) = max(a,c) = a
and
max(a, max(b, c)) = max(a, b) = a
Therefore q is true.
So p1  q is true.

67. 11. Existence Proofs Prove statements like
e.g. is there a British person in this classroom?
There are two main approaches:
1. Constructive existence proofs
2. Nonconstructive existence proofs
Does an element exist?

68. Constructive Existence Proof
Proof by Example
Find an element c, for which P(c) is true.
Then is true by Existential Generalization (EG). U P a b
... c d
...
some elements are in P;
others are outside

69. Example
Prove: “There is a positive integer that can be written as the sum of cubes of positive integers in two different ways.”
Proof: because
= =

70. Nonconstructive Existence Proof
Instead of proving x p(x) is true, we show that x p(x) is false
Translate: prove x p(x) is false
In words, we are trying to show that p() is not empty. U P a b
... c d
...
show that P is not empty
i.e. that is has some elements

71. 12. Counter-examples
A common problem is to show that a x P(x) statement is false:
e.g. prove “All the planets in our solar system are made of cheese.” is false
Logically:
prove x P(x) is false   x P(x) is true  x P(x) is true
In words, found an example that shows that P() is false – called a counter-example.
show that this
area is not empty U P a b
... c

72. Examples
“All the planets in our solar system are made of cheese.” Counter-example: the Earth. So the statement is false.
“Every positive integer is the sum of the squares of 3 integers.” Counter-example: 7. So the statement is false.

73. 13. More Information
Discrete Mathematics and its Applications Kenneth H. Rosen McGraw Hill, 2007, 7th edition
chapter 1, sections 1.6 – 1.8