1. Discrete Structures CSC 281
Logic Module – Part I (Propositional Logic)
Review

2. Conditional propositions and logical equivalence
A conditional proposition is of the form
“If p then q”
In symbols: p  q
Example:
p = " John is a programmer"
q = " Mary is a lawyer "
p  q = “If John is a programmer then Mary is a lawyer"
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3. p  q is true when both p and q are true
Truth table of p  q p q
p  q T F
or when p is false
p  q is true when both p and q are true
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4. Hypothesis and conclusion
In a conditional proposition p  q,
p is called the antecedent or hypothesis
q is called the consequent or conclusion
If "p then q" is considered logically the same as "p only if q"
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5. Necessary and sufficient
A necessary condition is expressed by the conclusion.
A sufficient condition is expressed by the hypothesis.
Example:
If John is a programmer then Mary is a lawyer"
Necessary condition: “Mary is a lawyer”
Sufficient condition: “John is a programmer”
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6. Logical equivalence p q ~p  q p  q T F
Two propositions are said to be logically
equivalent if their truth tables are identical.
Example: ~p  q is logically equivalent to p  q p q
~p  q
p  q T F
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7. Converse p q p  q q  p T F The converse of p  q is q  p
These two propositions
are not logically equivalent p q
p  q
q  p T F
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8. They are logically equivalent.
Contrapositive
The contrapositive of the proposition p  q is ~q  ~p.
They are logically equivalent. p q
p  q
~q  ~p T F
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9. p  q is logically equivalent to (p  q)^(q  p)
Double implication
The double implication “p if and only if q” is defined in symbols as p  q
p  q is logically equivalent to (p  q)^(q  p) p q
p  q
(p  q) ^ (q  p) T F
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10. Tautology
A proposition is a tautology if its truth table contains only true values for every case
Example: p  p v q p q
p  p v q T F
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11. Contradiction p p ^ (~p) T F
A proposition is a tautology if its truth table contains only false values for every case
Example: p ^ ~p p
p ^ (~p) T F
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12. De Morgan’s laws for logic
The following pairs of propositions are logically equivalent:
~ (p  q) and (~p)^(~q)
~ (p ^ q) and (~p)  (~q)
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13. Quantifiers
A propositional function P(x) is a statement involving a variable x
For example:
P(x): 2x is an even integer
x is an element of a set D
For example, x is an element of the set of integers
D is called the domain of P(x)
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14. Domain of a propositional function
In the propositional function
P(x): “2x is an even integer”,
the domain D of P(x) must be defined, for
instance D = {integers}.
D is the set where the x's come from.
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15. For every and for some
Most statements in mathematics and computer science use terms such as for every and for some.
For example:
For every triangle T, the sum of the angles of T is 180 degrees.
For every integer n, n is less than p, for some prime number p.
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16. Universal quantifier One can write P(x) for every x in a domain D
In symbols: x P(x)
 is called the universal quantifier
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17. Truth of as propositional function
The statement x P(x) is
True if P(x) is true for every x  D
False if P(x) is not true for some x  D
Example: Let P(n) be the propositional function n2 + 2n is an odd integer
n  D = {all integers}
P(n) is true only when n is an odd integer, false if n is an even integer.
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18. Existential quantifier
For some x  D, P(x) is true if there exists
an element x in the domain D for which P(x) is
true. In symbols: x, P(x)
The symbol  is called the existential quantifier.
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19. Counterexample
The universal statement x P(x) is false if x  D such that P(x) is false.
The value x that makes P(x) false is called a counterexample to the statement x P(x).
Example: P(x) = "every x is a prime number", for every integer x.
But if x = 4 (an integer) this x is not a primer number. Then 4 is a counterexample to P(x) being true.
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20. Generalized De Morgan’s laws for Logic
If P(x) is a propositional function, then each pair of propositions in a) and b) below have the same truth values:
a) ~(x P(x)) and x: ~P(x)
"It is not true that for every x, P(x) holds" is equivalent to "There exists an x for which P(x) is not true"
b) ~(x P(x)) and x: ~P(x)
"It is not true that there exists an x for which P(x) is true" is equivalent to "For all x, P(x) is not true"
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21. Summary of propositional logic
In order to prove the universally quantified statement x P(x) is true
It is not enough to show P(x) true for some x  D
You must show P(x) is true for every x  D
In order to prove the universally quantified statement x P(x) is false
It is enough to exhibit some x  D for which P(x) is false
This x is called the counterexample to the statement x P(x) is true

22. Discrete Structures CSC 281
Logic Module – Part II (proof methods)

23. Acknowledgement
Most of these slides were either created by Professor Bart Selman at Cornell University or else are modifications of his slides

24. Methods for Proving Theorems

25. Theorems, proofs, and Rules of Inference
When is a mathematical argument correct?
What techniques can we use to construct a mathematical argument?
Theorem – is a true proposition that is guaranteed by a proof.
Axioms or postulates – statements which are given and assumed to be true.
Proof – sequence of statements, a valid argument, to show that a theorem is true.
Rules of Inference – rules used in a proof to draw conclusions from assertions known to be true.
Note:
Lemma is a small theorem which is used to prove a bigger theorem.
A corollary is a theorem that can be proven to be a logical consequence of another theorem.
Conjecture is a statement believed to be true but for which there is not a proof yet. If the conjecture is proved true it becomes a thereom.
Fermat’s theorem was a conjecture for a long time.

26. Rules of Inference

27. Propositional logic: Rules of Inference or Methods of Proof
How to produce additional wffs (sentences) from other ones? What steps can we
perform to show that a conclusion follows logically from a set of hypotheses?
Example
Modus Ponens P
P  Q
______________
 Q
The hypotheses (premises) are written in a column and the conclusions below the bar
The symbol  denotes “therefore”. Given the hypotheses, the conclusion follows.
The basis for this rule of inference is the tautology (P  (P  Q))  Q)
[aside: check tautology with truth table to make sure]
In words: when P and P  Q are True, then Q must be True also. (meaning of
second implication)

28. Propositional logic: Rules of Inference or Methods of Proof
Example: Modus Ponens
If you study the CSC281 material  You will pass
You study the CSC281 material
______________
 you will pass
Nothing “deep”, but again remember the formal reason is that
((P ^ (P  Q))  Q is a tautology.

29. Propositional logic: Rules of Inference
See Table 1, p. 66, Rosen.
Rule of Inference
Tautology (Deduction Theorem)
Name P
 P  Q
P  (P  Q)
Addition
P  Q
 P
(P  Q)  P
Simplification Q
 P  Q
[(P)  (Q)]  (P  Q)
Conjunction
PQ
 Q
[(P)  (P Q)]  P
Modus Ponens
 Q
P  Q
 P
[(Q)  (P Q)]  P
Modus Tollens
Q  R
 P R
[(PQ)  (Q  R)]  (PR)
Hypothetical Syllogism
(“chaining”)
P  Q P
[(P  Q)  (P)]  Q
Disjunctive syllogism
P  R
 Q  R
[(P  Q)  (P  R)]  (Q  R)
Resolution

30. Rules of inference for quantified statements
1. Universal instantiation
 xD, P(x)
d  D
Therefore P(d)
2. Universal generalization
P(d) for any d  D
Therefore x, P(x)
3. Existential instantiation
 x  D, P(x)
Therefore P(d) for some d D
4. Existential generalization
P(d) for some d D
Therefore  x, P(x)
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31. Valid Arguments Show that (p1  p2  …pn)  q is a tautology
An argument is a sequence of propositions. The final proposition is called the conclusion of the argument while the other proposition are called the premises or hypotheses of the argument.
An argument is valid whenever the truth of all its premises implies the truth of its conclusion.
How to show that q logically follows from the hypotheses (p1  p2  …pn)?
Show that
(p1  p2  …pn)  q is a tautology
One can use the rules of inference to show the validity of an argument.

32. Proof Tree
Proofs can also be based on partial orders – we can represent them using a tree structure:
Each node in the proof tree is labeled by a wff, corresponding to a wff in the original set of hypotheses or be inferable from its parents in the tree using one of the rules of inference;
The labeled tree is a proof of the label of the root node.
Example:
Given the set of wffs:
P, R, PQ
Give a proof of Q  R

33. What rules of inference
Tree Proof
P, P Q, Q, R, Q  R
P PQ R Q
Q  R MP
Conj.
What rules of inference
did we use?

34. Length of Proofs Consider premises:
Why bother with inference rules? We could always use a truth table
to check the validity of a conclusion from a set of premises.
But, resulting proof can be much shorter than truth table method.
Consider premises:
p_1, p_1  p_2, p_2  p_3, …, p_(n-1)  p_n
To prove conclusion: p_n
Inference rules: Truth table:
n-1 MP steps 2n
Key open question: Is there always a short proof for any valid
conclusion? Probably not. The NP vs. co-NP question.
(The closely related: P vs. NP question carries a $1M prize.)

35. Valid Arguments Show that (p1  p2  …pn)  q is a tautology
An argument is a sequence of propositions. The final proposition is called the conclusion of the argument while the other propositions are called the premises or hypotheses of the argument.
An argument is valid whenever the truth of all its premises implies the truth of its conclusion.
How to show that q logically follows from the hypotheses (p1  p2  …pn)?
Show that
(p1  p2  …pn)  q is a tautology
One can use the rules of inference to show the validity of an argument.
Vacuous proof - if one of the premises is false then (p1  p2  …pn)  q
is vacuously True, since False implies anything.

36. Arguments involving universally quantified variables
Note: Many theorems involve statements for universally quantified variables:
e.g., the following statements are equivalent:
“If x>y, where x and y are positive real numbers, then x2 > y2 ”
“xy (if x > y > 0 then x2 > y2) ”
Quite often, when it is clear from the context, theorems are proved without explicitly using the laws of universal instantiation and universal generalization.

37. Methods of Proof Direct Proof Proof by Contraposition
Proof by Contradiction
Proof of Equivalences
Proof by Cases
Exhaustive Proof
Existence Proofs
Uniqueness Proofs
Counterexamples

38. Direct Proof
Proof of a statement p  q
Assume p
From p derive q.

39. ((M  C)  (D  C)  (D  S)  (M))  S
Example - direct proof
Here’s what you know:
Premises:
Mary is a Math major or a CS major.
If Mary does not like discrete math, she is not a CS major.
If Mary likes discrete math, she is smart.
Mary is not a math major.
Can you conclude Mary is smart?
Let
M - Mary is a Math major
C – Mary is a CS major
D – Mary likes discrete math
S – Mary is smart
Informally, what’s the chain of reasoning?
((M  C)  (D  C)  (D  S)  (M))  S ?

40. ((M  C)  (D  C)  (D  S)  (M))  S
Example - direct proof
In general, to prove p  q, assume p is true and show that q must also be true
Since, p is a conjunction of all the premises, we instead make the equivalent assumption that all of the following premises are true
M  C
D  C
D  S M
Then the truth of these premises are used to prove S is true
((M  C)  (D  C)  (D  S)  (M))  S ?
hese

41. Example - direct proof Mary is smart! QED
1. M  C Given
2. D  C Given
3. D  S Given
4. M Given
5. C
6. D
7. S
Disjunctive Syllogism (1,4)
Modus Tollens (2,5)
Modus Ponens (3,6)
Mary is smart!
QED
QED or Q.E.D. --- quod erat demonstrandum
“which was to be demonstrated”
or “I rest my case” 

42. Example 2: Direct Proof Theorem: If n is odd integer, then n2 is odd.
Two definitions:
The integer is even if there exists an integer k such that n = 2k.
An is odd if there exists an integer k such that n = 2k+1.
Note: An integer is either even or odd, but not both.
This is an immediate consequence of the division algorithm: If a and b are positive integers, then there exist unique integers q and r with a = qb + r and 0  r < b
Other proofs can also be given, depending on what previous facts have already been established.
This fact is not needed in the first proof, is needed in a later proofs.
n (n is odd)  (n2 is odd) E

43. Example 2: Direct Proof Theorem: (n) P(n)  Q(n),
where P(n) is “n is an odd integer” and Q(n) is “n2 is odd.”
We will show P(n)  Q(n)

44. Example 2: Direct Proof
Theorem:
If n is odd integer, then n2 is odd.
Proof:
Let p denote “n is odd integer” and q denote “n2 is odd”; we want to show that p  q
Assume p, i.e., n is odd.
By definition n = 2k + 1, where k is some integer.
Therefore n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2 (2k2 + 2k ) + 1, which is by
definition an odd number (k’ = (2k2 + 2k ) ).
QED
Proof strategy hint: Go back to definitions of concepts
and start by trying a direct proof.

45. Proof by Contraposition
Proof of a statement p  q by contraposition
Recall the tautology of the equivalence of a implication and its contrapositive.
p  q  q  p (the contrapositive)
So, we can prove p  q by establishing the equivalent statement that
¬q  ¬ p
So, we prove the implication p  q by first assuming q, and showing that p follows
raposi

46. Example 1: Proof by Contraposition
Example: Prove that if a and b are integers, and a + b ≥ 15, then a ≥ 8 or b ≥ 8.
(a + b ≥ 15)  (a ≥ 8) v (b ≥ 8)
Proof strategy:
Note that negation
of conclusion is
easier to start with
here.
(Assume q) Suppose (a < 8)  (b < 8).
(Show p) Then (a ≤ 7)  (b ≤ 7),
and (a + b) ≤ 14,
and (a + b) < 15.
QED

47. Example 2: Proof by Contraposition
Theorem
For an integer n,
if 3n + 2 is odd, then n is odd.
I.e. For n integer,
3n+2 is odd  n is odd
Proof by Contraposition:
Let p denote “3n + 2” is odd and q denote “n is odd”; we must show that p  q
The contraposition of our theorem is ¬q  ¬p
n is even  3n + 2 is even
Now we can use a direct proof
Assume ¬q , i.e, n is even therefore n = 2 k for some k
Therefore 3 n + 2 = 3 (2k) + 2 = 6 k + 2 = 2 (3k + 1) which is even.
QED
Again, negation
of conclusion is
easy to start with.
Try direct proof. 

48. Proof by Contradiction
A – We want to prove p.
We show that:
¬p  F (i.e., a False statement , say r ¬r)
We conclude that ¬p is false since (1) is True and therefore p is True.
B – We want to show p  q
Assume the negation of the conclusion, i.e., ¬q
Show that (p  ¬q )  F
Since ((p  ¬q )  F)  (p  q) (why?) we are done
((p  ¬q )  F)  (p  ¬q )
 p  q

49. Example 1: Proof by Contradiction
Rainy days make gardens grow.
Gardens don’t grow if it is not hot.
When it is cold outside, it rains.
Prove that it’s hot.
Hmm. We will assume “not Hot” ≡ “Cold”
Let
R – Rainy day
G – Garden grows
H – It is hot
Given: R  G
H  G
H  R
Show: H
((R  G)  (H  G)  (H  R))  H ?

50. Example 1: Proof by Contradiction
Aside: we assume it’s either Hot or it is not Hot.
Called the “law of excluded middle”. In certain complex
arguments, it’s not so clearly valid. (hmm…) This led to
“constructive mathematics” and “intuitionistic mathematics”.
Given: R  G
H  G
H  R
Show: H
1. R  G Given
2. H  G Given
3. H  R Given
4. H assume negation of conclusion at
5. R MP (3,4)
6. G MP (1,5)
7. G MP (2,4)
8. G  G contradiction
 H

51. Example2: Proof by Contradiction
Classic proof that 2 is irrational.
Suppose 2 is rational. Then 2 = a/b for some integers a and b (relatively prime; no factor in common).
It’s quite clever!!
Note: Here we again first go to the definition of concepts (“rational”). Makes sense! Definitions provide
information about important concepts. In a sense, math is all about “What follows from the definitions and premises!
2 = a/b implies
2 = a2/b2
ion
2b2 = a2
a2 is even, and so a is even (a = 2k for some k)
But if a and b are both even, then they are not relatively prime!
Q.E.D.
2b2 = (2k)2 = 4k2
b2 = 2k2
b2 is even, and so b is even (b = 2k for some k)

52. Example2: Proof by Contradiction
You’re going to let me get away with that? 
Lemma: a2 is even implies that a is even (i.e., a = 2k for some k)??
Suppose to the contrary that a is not even.
Then a = 2k + 1 for some integer k
Then a2 = (2k + 1)(2k + 1) = 4k2 + 4k + 1
and a2 is odd.
Then, as discussed earlier, a2 is not even
contradiction
So, a really is even.
Corollary: An integer n is even if and only if n2 is even
Why does the above statement follow immediately from previous work???

53. Example 3: Proof by Contradiction
Theorem:
“There are infinitely many prime numbers”
Proof by contradiction
Let P – “There are infinitely many primes”
Assume ¬P, i.e., “there is a finite number of primes” , call largest p_r.
Let’s define R the product of all the primes, i.e, R = p_1 × p_2 × … × p_r.
Consider R + 1.
Now, R+1 is either prime or not:
If it’s prime, we have prime larger than p_r.
If it’s not prime, let p* be a prime dividing (R+1). But p* cannot be any of p_1, p_2, … p_r (remainder 1 after division); so, p* not among initial list and thus p* is larger than p_r.
This contradicts our assumption that there is a finite set of primes, and therefore such an assumption has to be false which means that there are infinitely many primes.
(Euclid’s proof, c 300 BC)
One of the most famous
early proofs. An early
intellectual “tour the force”.
(Clever “trick”. The key to the proof.)
For more details, see Or
Also, non-constructive.
See e.g.

54. Example 4: Proof by Contradiction
Theorem “If 3n+2 is odd, then n is odd”
Let p = “3n+2 is odd” and q = “n is odd”
1 – assume p and ¬q i.e., 3n+2 is odd and n is not odd
2 – because n is not odd, it is even
3 – if n is even, n = 2k for some k, and therefore 3n+2 = 3 (2k) + 2 = 2 (3k + 1), which is even
4 – So, we have a contradiction, 3n+2 is odd and 3n+2 is even.
Therefore, we conclude p  q, i.e., “If 3n+2 is odd, then n is odd”
Q.E.D.

55. Proof of Equivalences To prove p  q show that p q and q p.
The validity of this proof results from the fact that
(p  q)  [ (p q)  (q p)] is a tautology

56. Counterexamples Show that (x) P(x) is false
We need only to find a counterexample.

57. Counterexample  Show that the following statement is false:
“Every day of the week is a weekday”
Proof:
Saturday and Sunday are weekend days. 

58. Proof by Cases To show (p1  p2 … pn )  q We use the tautology
[(p1  p2 … pn )  q ]  [(p1  q )  (p2  q) … (pn  q )]
A particular case of a proof by cases is an exhaustive proof in which all the cases are considered

59. Theorem
“If n is an integer, then n2 ≥ n ”
Proof by cases
Case 1 n=0 02 = 0
Case 2 n > 0, i.e., n  1. We get n2 ≥ n since we can multiply both sides of the inequality by n, which is positive.
Case 3 n < 0. Then nn > 0n since n is negative and multiplying both sides of inequality by n changes the direction of the inequality). So, we have n2 > 0 in this case.
In conclusion, n2 ≥ n since this is true in all cases.

60. Existence Proofs Existence Proofs: Constructive existence proofs
Example: “there is a positive integer that is the sum of cubes of positive integers in two different ways”
Proof: Show by brute force using a computer 1729 = =
Non-constructive existence proofs
Example: “n (integers), p so that p is prime, and p > n.”
Proof: Recall proof used to show there were infinitely many primes.
Very subtle – does not give an example of such a number, but shows one exists. (Let P = product of all primes < n and consider P+1. )
Uniqueness proofs involve
Existence proof
Uniqueness proof

61. Example 1 - Existence Proofs
NON-CONSTRUCTIVE
n (integers), p so that p is prime, and p > n.
Proof: Let n be an arbitrary integer, and consider n! If (n! + 1) is prime, we are done since (n! + 1) > n. But what if (n! + 1) is composite?
If (n! + 1) is composite then it has a prime factorization, p1p2…pn = (n! + 1)
Consider the smallest pi, and call it p. How small can it be?
Can it be 2?
Can it be 3?
Can it be 4?
So, p > n, and we are done. BUT WE DON’T KNOW WHAT p IS!!!
Can it be n?

62. Example 2: Existence proof
Thm. There exists irrational numbers x and y such that xy is rational.
Proof.
2 is irrational (see earlier proof).
Consider: z = 22
We have two possible cases:
z is rational. Then, we’re done (take x = 2 and y = 2 ).
z is irrational. Now, let x = z and y = 2. And consider:
xy = (22 )2 = 2(2 2) = 22 = 2 , which is rational. So, we’re done.
Since, either 1) or 2) must be true, it follows that there does exist irrational x
and y such that xy is rational. Q.E.D.
“Start with something you know
about rational / irrational numbers.”
Non-constructive!

63. Example 3: Non-constructive proof
Poisonous
From game theory.
Consider the game “Chomp”.
Two players. Players take turn eating
at least one of the remaining cookies.
At each turn, the player also eats all cookies to the left and below the cookie he or she selects.
The player who is “forced” to eat the poisened cookie loses. 
Is there a winning strategy for either player?
m x n cookies
Winning strategy for a player:
“A way of making moves” that is guaranteed
to lead to a win, no matter what the
opponent does. (How big to write down?)

64. Is first choice of the bottom right cookie essential? If so, why?
Claim: First player has a winning strategy!
Proof. (non-constructive)
First, note that the game cannot end in a “draw”. After at most m x n moves, someone has eaten the last cookie. 
Consider the following strategy for the first player:
--- Start by eating the cookie in the bottom right corner.
--- Now, two possibilities:
This is part of a winning strategy for 1st player (and thus player has winning strategy). OR
2nd player can now make a move
that is part of the winning strategy for the 2nd player.
But, if 2) is the case, then 1st player can follow
a winning strategy by on the first move making
the move of the second player and following his
or her winning strategy! So, again, 1st player has
winning strategy. Q.E.D.
Three possible moves
This is called a “strategy
stealing” argument. Think
through carefully to convince
yourself!
(Actual strategy not known for
general boards!)
Corner is “null move”
Is first choice of the bottom right cookie essential? If so, why?

65. Fallacies Fallacies are incorrect inferences. Some common fallacies:
The Fallacy of Affirming the Consequent
The Fallacy of Denying the Antecedent
Begging the question or circular reasoning

66. The Fallacy of Affirming the Consequent
If the butler did it he has blood on his hands.
The butler had blood on his hands.
Therefore, the butler did it.
This argument has the form
PQ Q
 P
or ((PQ)  Q)P which is not a tautology and therefore not a valid rule of inference

67. The Fallacy of Denying the Antecedent
If the butler is nervous, he did it.
The butler is really mellow.
Therefore, the butler didn't do it.
This argument has the form
PQ ¬P
 ¬Q
or ((PQ)  ¬P) ¬Q which is not a tautology and therefore not a valid rule of inference

68. Begging the question or circular reasoning
This occurs when we use the truth of the statement being proved (or
something equivalent) in the proof itself.
Example:
Conjecture: if n2 is even then n is even.
Proof: If n2 is even then n2 = 2k for some k. Let n = 2m for some m. Hence, x must be even.
Note that the statement n = 2m is introduced without any argument showing it.

69. Final example Tiling X X Notoriously hard problem
automated theorem prover
--- requires “true cleverness”
Final example Tiling
62 squares: 32 black
30 white
31 doms.: 31 black
31 white squares! X
A domino
Can you use 32 dominos to
cover the board?
Easily!
(many ways!)
What about the mutilated
checkerboard? Hmm…
Use counting?
No! Why?
What is the proof based upon?
Proof uses clever coloring
and counting argument.
Note: also valid for board
and dominos without b&w pattern!
(use proof by contradiction) X
Standard checkerboard. 8x8 = 64 squares

70. Additional Proof Methods Covered in CSC281
Induction Proofs
Combinatorial proofs
But first we have to cover some basic notions on sets, functions, and counting.