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10-2
Discrete Mathematics – Theorems and Proofs
Previous Lecture
Axioms and theorems
Rules of inference for quantified statements
Rule of universal specification
Rule of universal generalization
Existential rules
Theorems and Proofs II
Introduction
Discrete Mathematics
Andrei Bulatov
Discrete Mathematics – Theorems and Proofs
10-3
Methods of Proving – Direct Proofs
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Discrete Mathematics – Theorems and Proofs
Methods of Proving – Direct Proofs
Direct proofs are used when we need to proof statements like
∀x (P(x) → Q(x))
Example: ``If 2x – 6 = 0 then x = 3.’’
Notation: P(x) - ``2x – 6 = 0’’, Q(x) - ``2x = 6’’, R(x) - ``x = 3’’
Main steps
Our goal is to prove that P(a) → Q(a) is a tautology for a generic
value a.
Need to prove: ∀x (P(x) → R(x))
Previous knowledge: ∀x (P(x) → Q(x)), ∀x (Q(x) → R(x))
1. Assume that P(a) is true
Step
Reason
1. P(c)
assumption
1. ∀x (P(x) → Q(x)), ∀x (Q(x) → R(x)) premises
2. Using axioms, previous theorems etc. prove that Q(a) is true
3. Conclude that P(a) → Q(a) is true
4. Use the rule of universal generalization to infer
∀x (P(x) → Q(x))
Discrete Mathematics – Theorems and Proofs
Example
2. P(c) → Q(c), Q(c) → R(c),
rule of univ. specification
3. R(c)
4. ∀x (P(x) → R(x))
Modus Ponens
rule of univ. generalization
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Discrete Mathematics – Theorems and Proofs
Example
Theorem. Everyone who studies more knows less.
Proof. Everyone who studies more knows less.
S(x) x studies more
M(x) x knows more
F(x) x forgets more
L(x) x knows less
Premises: ∀x (S(x) → M(x)), ∀x (M(x) → F(x)), ∀x (F(x) → L(x))
Theorem: ∀x (S(x) → L(x))
Step
Reason
1. S(c)
assumption
2. ∀x (S(x) → M(x)), ∀x (M(x) → F(x)), ∀x (F(x) → L(x)) premises
3. S(c) → M(c), M(c) → F(c), F(c) → L(c)
rule of univ. spec.
4. L(c)
Modus Ponens
5. ∀x (S(x) → L(x))
rule of univ. gen.
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Methods of Proving – Proof by Contraposition
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Methods of Proving – Proof by Contraposition (cntd)
Sometimes direct proofs do not work
So assume that n is odd, that is there is k such that n = 2k + 1.
Definition: n is even if and only if there is k such that n = 2k
Then 3n + 2 = 3⋅(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1.
That is 3n + 2 is odd.
Prove that if 3n + 2 is even, then n is also even
That is ∀x (E(3x + 2) → E(x))
We have proved that ¬E(3n + 2) is true, and therefore the
contraposition ∀x (¬E(x ) → ¬E(3x + 2)) is true.
Finally, we conclude that the theorem ∀x (E(3x + 2) → E(x)) is also
true.
Let us try the direct approach:
As for the generic value n the number 3n + 2 is even, for some
k we have 3n + 2 = 2k. Therefore 3n = 2(k + 1).
Now what?
What if instead of ∀x (E(3x + 2 ) → E(x)) we prove the
contrapositive, ∀x (¬E(x ) → ¬E(3x + 2)) ?
Discrete Mathematics – Theorems and Proofs
10-9
Methods of Proving – Proof by Contraposition (cntd)
Proofs by contradiction use the Rule of Contradiction
¬p → F
∴p
Can be used to prove statements of any form
1. Assume that ¬Q(a) is true
Main steps
1. Assume ¬p.
2. Using axioms, previous theorems etc. infer a contradiction
3. Conclude p.
2. Using axioms, previous theorems etc. prove that ¬P(a) is true
3. Conclude that ¬Q(a) → ¬P(a) is true
4. Conclude that P(a) → Q(a) is true
5. Use the rule of universal generalization to infer
Usually the contradiction has the form ∃x (Q(x) ∧ ¬Q(x))
∀x (P(x) → Q(x))
Example
Definition: a barber is called strict if he shaves those and only
those who do not shave themselves.
Theorem. There is no strict barber.
(All barbers are not strict.)
Proof.
Assume the contrary: a strict barber c exists
Does he shave himself?
If no (¬q), then by the definition he must shave himself (q)
If yes (q), then by definition he must not (¬q)
Either way we have q ∧ ¬q, a contradiction
We conclude that a strict barber does not exist
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Methods of Proving – Proof by Contradiction
Main steps
Our goal is to prove that P(a) → Q(a) is a tautology for a generic
value a.
Instead we prove the contrapositive ¬Q(a) → ¬P(a)
Discrete Mathematics – Theorems and Proofs
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Example (cntd)
© Springer
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Discrete Mathematics – Theorems and Proofs
Pythagoras
Another Example
Definition: a real number is said to be rational if it can be
represented as a fraction
Prove that
Proof
Suppose that
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Discrete Mathematics – Theorems and Proofs
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a
where a,b are integers
b
c
2
c
b
2 is irrational
b
a
2
2 is rational, that is there are integers a,b such
a
2= .
b
We may assume that a,b have no common divisor.
2
2
Squaring we obtain a = 2b .
2
Since a is even, a is also even, hence a = 2c for some c.
2
2
2
2
Therefore 2b = 4c , and so b = 2c .
that
2
c2 = a 2 + b2
``Number is the ruler of forms and ideas, and the
cause of gods and demons’’
numbers = rational numbers
2 does not belong to this world
Hence b is even.
We get that a and b have a common factor – 2. A contradiction.
Discrete Mathematics – Theorems and Proofs
a
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Proving Existential Statements
How to prove ∃x P(x).
Constructive proofs: find or construct a value a such that P(a) is
true.
Prove that there is a grey car…
My car is grey!
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Homework
Exercises from the Book:
No. 5, 9, 11, 13, 15, 17 (page 116-117)
Pure proofs of existence:
Assume that ∀x ¬P(x).
Using axioms, previous theorems etc. infer a contradiction
Thus, this is a proof by contradiction.
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