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Discrete Structures
CSC 281
Logic Module – Part I (Propositional Logic)
Review
1
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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Truth table of p  q
p
q
pq
T
T
T
T
F
F
F
T
T
F
F
T
or when p is false
 p  q is true when both p and q are true
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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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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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Logical equivalence
 Two propositions are said to be logically
equivalent if their truth tables are identical.
p
q
~p  q
pq
T
T
T
F
T
F
T
F
F
F
T
F
T
T
T
T
 Example: ~p  q is logically equivalent to p  q
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Converse
The converse of p  q is q  p
p
T
q
T
pq
T
qp
T
T
F
F
F
T
F
F
T
T
T
F
T
These two propositions
are not logically equivalent
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Contrapositive
The contrapositive of the proposition p  q is ~q  ~p.
p
T
q
T
pq
T
~q  ~p
T
T
F
F
T
F
T
F
T
F
F
T
T
They are logically equivalent.
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Double implication
The double implication “p if and only if q” is defined in symbols as p 
q
p
q
pq
(p  q) ^ (q  p)
T
T
T
T
T
F
F
F
F
T
F
F
F
F
T
T
p  q is logically equivalent to (p  q)^(q  p)
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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pvq
T
T
T
T
F
T
F
T
T
F
F
T
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Contradiction
A proposition is a tautology if its truth table contains only false
values for every case
– Example: p ^ ~p
p
p ^ (~p)
T
F
F
F
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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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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)
13
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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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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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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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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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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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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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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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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
Discrete Structures
CSC 281
Logic Module – Part II (proof methods)
22
Acknowledgement
Most of these slides were either created by
Professor Bart Selman at Cornell University or
else are modifications of his slides
23
Methods for Proving Theorems
24
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.
o Fermat’s theorem was a conjecture for a long time.
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Rules of Inference
26
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)
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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.
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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
P
Q
PQ
[(P)  (Q)]  (P  Q)
Conjunction
P
PQ
Q
[(P)  (P Q)]  P
Modus Ponens
Q
PQ
 P
[(Q)  (P Q)]  P
Modus Tollens
PQ
QR
 P R
[(PQ)  (Q  R)]  (PR)
Hypothetical Syllogism
(“chaining”)
PQ
P
Q
[(P  Q)  (P)]  Q
Disjunctive syllogism
PQ
P  R
QR
[(P  Q)  (P  R)]  (Q  R)
Resolution
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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Valid Arguments
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.
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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
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Tree Proof
P, P Q, Q, R, Q  R
P
PQ
R
MP
Q
Conj.
QR
What rules of inference
did we use?
33
Length of Proofs
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: n-1 MP steps Truth table:
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.)
Valid Arguments
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.
35
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.
36
Methods of Proof
Direct Proof
Proof by Contraposition
Proof by Contradiction
Proof of Equivalences
Proof by Cases
Exhaustive Proof
Existence Proofs
Uniqueness Proofs
Counterexamples
37
Direct Proof
Proof of a statement
pq
Assume p
From p derive q.
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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.
Let
Mary is not a math major.
Can you conclude Mary is smart?
Informally, what’s the chain of reasoning?
M - Mary is a Math major
C – Mary is a CS major
D – Mary likes discrete math
S – Mary is smart
((M  C)  (D  C)  (D  S)  (M))  S
?
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Example - direct proof
In general, to prove p  q, assume p is true and show that q must also be
true
((M  C)  (D  C)  (D  S)  (M))  S
?
• 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
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Example - direct proof
1. M  C
2. D  C
3. D  S
4. M
Given
Given
Given
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” 
41
Example 2: Direct Proof
Theorem:
If n is odd integer, then n2 is odd.
n (n is odd)  (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.
42
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)
43
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.
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
45
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)
(Assume q)
(Show p)
Suppose (a < 8)  (b < 8).
Then (a ≤ 7)  (b ≤ 7),
and (a + b) ≤ 14,
and (a + b) < 15.
Proof strategy:
Note that negation
of conclusion is
easier to start with
here.
QED
46
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
Again, negation
of conclusion is
easy to start with.
Try direct proof. 
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
Proof by Contradiction
A – We want to prove p.
We show that:
(1) ¬p  F (i.e., a False statement , say r ¬r)
(2) We conclude that ¬p is false since (1) is True and therefore p is True.
B – We want to show p  q
(1) Assume the negation of the conclusion, i.e., ¬q
(2) Show that (p  ¬q )  F
(3) Since ((p  ¬q )  F)  (p  q) (why?) we are done
((p  ¬q )  F)  (p  ¬q )
 p q
48
Example 1: Proof by Contradiction
Example:
Hmm. We will assume “not Hot” ≡ “Cold”
Rainy days make gardens grow.
Gardens don’t grow if it is not hot.
When it is cold outside, it rains.
Let
R – Rainy day
G – Garden grows
H – It is hot
Prove that it’s hot.
Given: R  G
H  G
H  R
Show: H
((R  G)  (H  G)  (H  R))  H
?
49
Example 1: Proof by Contradiction
Given: R  G
H  G
H  R
Show: H
1. R  G
2. H  G
3. H  R
4. H
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
Given
Given
assume negation of conclusion
5. R
6. G
MP (3,4)
MP (1,5)
7. G
8. G  G
MP (2,4)
contradiction
H
50
Example2: Proof by Contradiction
Classic proof that 2 is irrational.
It’s quite clever!!
Suppose 2 is rational. Then 2 = a/b for some integers a and b
(relatively prime; no factor in common).
Note: Here we again first go to the
definition of concepts (“rational”).
Makes sense! Definitions provide
2 = a/b implies
information about important concepts.
2 = a2/b2
In a sense, math is all about “What
2b2 = a2
follows from the definitions and
premises!
a2 is even, and so a is even (a = 2k for some k)
2b2 = (2k)2 = 4k2
b2 = 2k2
b2 is even, and so b is even (b = 2k for some k)
But if a and b are both
even, then they are not
relatively prime!
Q.E.D.
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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
So, a really is even.
contradiction
Corollary: An integer n is even if and only if n2 is even
Why does the above statement follow immediately from previous work???
52
Example 3: Proof
Theorem:
“There are infinitely many prime numbers”
by Contradiction
(Euclid’s proof, c 300 BC)
One of the most famous
early proofs. An early
Proof by contradiction
intellectual “tour the force”.
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.
(Clever “trick”. The key to the proof.)
• 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.
Also, non-constructive.
See e.g. http://odin.mdacc.tmc.edu/~krc/numbers/infitude.html
http://primes.utm.edu/notes/proofs/infinite/euclids.html
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.
54
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
55
Counterexamples
Show that (x) P(x) is false
We need only to find a counterexample.
56
Counterexample
Show that the following statement is false:
“Every day of the week is a weekday”
Proof:
Saturday and Sunday are weekend days.

57
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
58
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.
59
Existence Proofs
Existence Proofs:
• Constructive existence proofs
o Example: “there is a positive integer that is the sum of cubes of positive
integers in two different ways”
o Proof: Show by brute force using a computer 1729 = 103 + 93 = 123 + 13
• Non-constructive existence proofs
o Example: “n (integers), p so that p is prime, and p > n.”
o Proof: Recall proof used to show there were infinitely many primes.
o 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
60
NON-CONSTRUCTIVE
Example 1 -
Existence Proofs
n (integers), p so that p is prime, and p > n.
Proof: Let n be an arbitrary integer, and consider n! + 1. 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 KNOWCan it be n?
WHAT p IS!!!
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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
“Start with something you know
about rational / irrational numbers.”
Non-constructive!
We have two possible cases:
1) z is rational. Then, we’re done (take x = 2 and y = 2 ).
2) z is irrational. Now, let x = z and y = 2. And consider:
x y = ( 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.
Poisonous
Example 3: Non-constructive proof
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?)
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:
1) This is part of a winning strategy for 1st
player (and thus player has winning strategy).
OR
2) 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?
Fallacies
Fallacies are incorrect inferences. Some common fallacies:
1. The Fallacy of Affirming the Consequent
2. The Fallacy of Denying the Antecedent
3. Begging the question or circular reasoning
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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
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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
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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.
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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!
A domino
Can you use 32 dominos to
cover the board? Easily!
(many ways!)
What about the mutilated
checkerboard? Hmm… No! Why?
Use counting?
X
X
Standard checkerboard. 8x8 = 64 squares
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)
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Additional Proof Methods
Covered in CSC281
• Induction Proofs
• Combinatorial proofs
But first we have to cover some basic notions on sets,
functions, and counting.
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