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Discrete Structures
Introduction to Proofs
Dr. Muhammad Humayoun
Assistant Professor
COMSATS Institute of Computer Science, Lahore.
[email protected]
https://sites.google.com/a/ciitlahore.edu.pk/dstruct/
Some material is taken from Dr. Atif’s slides
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Terminology
• Theorem: a statement that can be shown true.
Sometimes called facts.
• Proof: Demonstration that a theorem is true.
• Axiom: A statement that is assumed to be true.
• Lemma: a less important theorem that is useful
to prove a theorem.
• Corollary: a theorem that can be proven directly
from a theorem that has been proved.
• Conjecture: a statement that is being proposed to
be a true statement.
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Stating Theorems
• Theorem. If 𝑥 > 𝑦, where x and y are positive
real numbers, then 𝑥 2 > 𝑦 2 .
• Theorem. For all positive real numbers x and
y, if 𝑥 > 𝑦, then 𝑥 2 > 𝑦 2 .
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Methods of Proving Theorems
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Trivial Proofs
• Consider an implication: 𝑝 → 𝑞
• If it can be shown that p is true, then the
implication is always true
– By definition of an implication
• Note that you are showing that the conclusion
is true
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Trivial Proof Example
Consider the statement:
• If you are in CSC102 then you are a student.
• Since all people in CSC102 are students, the
implication is true.
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Vacuous proofs
• Consider an implication: 𝑝 → 𝑞
• If it can be shown that 𝑝 is false, then the
implication is always true.
– By definition of an implication
• Note that you are showing that the hypothesis
is false
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Example
Consider the statement:
• Every student snooze during class when he is
tired.
• Rephrased: If a student is tired then he snoozes
during class
• Since there is no such student who is snoozing,
(the hypothesis is false), the implication is true
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Methods of Proving Theorems
Important ones
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Methods of Proving Theorems
Direct Proofs
• Consider an implication: 𝑝 → 𝑞
– If p is false, then the implication is always true
– Thus, show that if p is true, then q is true
• To perform a direct proof, assume that p is
true, and show that q must therefore be
true
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Example Direct Proof
Theorem. Show that the square of an even number
is an even number.
For every number 𝑛, if 𝑛 is even, then 𝑛2 is even.
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Example Direct Proof
Theorem. Show that the square of an even number
is an even number.
For every number 𝑛, if 𝑛 is even, then 𝑛2 is even.
∀𝒏[even 𝒏 → even 𝒏𝟐 ]
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Example Direct Proof
Theorem. Show that the square of an even number
is an even number.
For every number 𝑛, if 𝑛 is even, then 𝑛2 is even.
∀𝒏[even 𝒏 → even 𝒏𝟐 ]
Proof.
Usual convention: Universal instantiation is not
explicitly used and it is directly showed that
even(𝑛) implies even(𝑛2 ).
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Example Direct Proof
Theorem. Show that the square of an even number is
an even number.
For every number 𝑛, if 𝑛 is even, then 𝑛2 is even.
∀𝒏[even 𝒏 → even 𝒏𝟐 ]
Proof.
Usual convention: Universal instantiation is not explicitly used and it
is directly showed that even(𝑛) implies even(𝑛2 ).
Assume 𝑛 is even.
Thus, 𝑛 = 2𝑘, for some 𝑘 (definition of even
numbers).
𝑛2 = 2𝑘 2 = 4𝑘 2 = 2(2𝑘 2 )
As 𝑛2 is 2 times an integer, 𝑛2 is thus even.
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Example Direct Proof
Theorem. Show that the square of an even number is
an even number.
For every number 𝑛, if 𝑛 is even, then 𝑛2 is even.
∀𝒏[even 𝒏 → even 𝒏𝟐 ]
Proof.
Usual convention: Universal instantiation is not explicitly used and it
is directly showed that even(𝑛) implies even(𝑛2 ).
Assume 𝑛 is even.
Thus, 𝑛 = 2𝑘, for some 𝑘 (definition of even
numbers).
𝑛2 = 2𝑘 2 = 4𝑘 2 = 2(2𝑘 2 )
As 𝑛2 is 2 times an integer, 𝑛2 is thus even.
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Proof by Contraposition
(or Indirect proofs)
• 𝑝 → 𝑞 ≡ ¬𝑞 → ¬𝑝
– If the antecedent (¬𝑞) is false, then the
contrapositive is always true
– Thus, show that if ¬𝑞 is true, then ¬𝑝 is true
• To perform an indirect proof, do a direct proof
on the contrapositive
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Indirect proof example
Theorem. If 𝑛2 is an odd integer then 𝑛 is an odd integer
Proof (by contrapositive).
We show that the contrapositive of the theorem statement
holds, therefore the theorem statement hold.
Contrapositive: If 𝑛 is an even integer, then 𝑛2 is an even
integer.
𝑝 → 𝑞 ≡ ¬𝑞 → ¬𝑝
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Indirect proof example
Theorem. If 𝑛2 is an odd integer then 𝑛 is an odd integer
Proof (by contrapositive).
We show that the contrapositive of the theorem statement
holds, therefore the theorem statement hold.
Contrapositive: If 𝑛 is an even integer, then 𝑛2 is an even
integer.
Assume that 𝑛 is even. Then by the definition of even
numbers: 𝑛 = 2𝑘 for some integer 𝑘.
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Indirect proof example
Theorem. If 𝑛2 is an odd integer then 𝑛 is an odd integer
Proof (by contrapositive).
We show that the contrapositive of the theorem statement
holds, therefore the theorem statement hold.
Contrapositive: If 𝑛 is an even integer, then 𝑛2 is an even
integer.
Assume that 𝑛 is even. Then by the definition of even
numbers: 𝑛 = 2𝑘 for some integer 𝑘. Thus
𝑛2 = 2𝑘 2 = 4𝑘 2 = 2(2𝑘 2 )
Since 𝑛2 is 2 times an integer, it is even. Hence our proof by
contraposition succeeds.
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Selecting a Proof method
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then
𝑛 is even.
Proof. (Via direct proof). Assume that 𝑛3 + 5 is
odd.
Then by the definition of odd numbers 𝑛3 + 5 =
2𝑘 + 1 for some integer k. Thus
𝑛3 = 2𝑘 − 4
…
???
Direct proof doesn’t seem to work.
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Selecting a Proof method
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then
𝑛 is even.
Proof. (Via direct proof). Assume that 𝑛3 + 5 is
odd.
Then by the definition of odd numbers 𝑛3 + 5 =
2𝑘 + 1 for some integer k. Thus
𝑛3 = 2𝑘 − 4
…
???
Direct proof doesn’t seem to work.
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Selecting a Proof method
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then 𝑛 is even.
Indirect proof.
Contrapositive: If 𝑛 is odd, then 𝑛3 + 5 is even.
Assume 𝑛 is odd, and show that 𝑛3 + 5 is even.
By the definition of odd numbers 𝑛 = 2𝑘 + 1 for some
integer 𝑘.
𝑛3 + 5 = 2𝑘 + 1 3 + 5 = 8𝑘 3 + 12𝑘 2 + 6𝑘 + 6
= 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3)
As 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3) is 2 times an integer, it is
even
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Selecting a Proof method
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then 𝑛 is even.
Indirect proof.
Contrapositive: If 𝑛 is odd, then 𝑛3 + 5 is even.
Assume 𝑛 is odd, and show that 𝑛3 + 5 is even.
By the definition of odd numbers 𝑛 = 2𝑘 + 1 for some
integer 𝑘.
𝑛3 + 5 = 2𝑘 + 1 3 + 5 = 8𝑘 3 + 12𝑘 2 + 6𝑘 + 6
= 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3)
As 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3) is 2 times an integer, it is
even
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Selecting a Proof method
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then 𝑛 is even.
Indirect proof.
Contrapositive: If 𝑛 is odd, then 𝑛3 + 5 is even.
Assume 𝑛 is odd, and show that 𝑛3 + 5 is even.
By the definition of odd numbers 𝑛 = 2𝑘 + 1 for some
integer 𝑘.
𝑛3 + 5 = 2𝑘 + 1 3 + 5 = 8𝑘 3 + 12𝑘 2 + 6𝑘 + 6
= 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3)
As 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3) is 2 times an integer, it is
even
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Selecting a Proof method
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then 𝑛 is even.
Indirect proof.
Contrapositive: If 𝑛 is odd, then 𝑛3 + 5 is even.
Assume 𝑛 is odd, and show that 𝑛3 + 5 is even.
By the definition of odd numbers 𝑛 = 2𝑘 + 1 for some
integer 𝑘.
𝑛3 + 5 = 2𝑘 + 1 3 + 5 = 8𝑘 3 + 12𝑘 2 + 6𝑘 + 6
= 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3)
As 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3) is 2 times an integer, it is
even
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Proof by contradiction
(another type of indirect proofs)
• Given a statement of the form 𝑝 → 𝑞
• Assume p is true and q is false
– Assume p and assume ¬𝑞
• Then prove that ¬𝑞 cannot occur
– A contradiction exists
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Proof by contradiction example
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then 𝑛 is even.
Rephrased: If 𝑛3 + 5 is odd, then 𝑛 is even
Proof. Assume p is true and q is false (Assume 𝑝. Assume ¬𝑞)
Assume that 𝑛3 + 5 is odd, and 𝑛 is odd.
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Proof by contradiction example
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then 𝑛 is even.
Rephrased: If 𝑛3 + 5 is odd, then 𝑛 is even
Proof. Assume p is true and q is false (Assume 𝑝. Assume ¬𝑞)
Assume that 𝑛3 + 5 is odd, and 𝑛 is odd.
By the definition of odd numbers 𝑛 = 2𝑘 + 1 for some
integer k.
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Proof by contradiction example
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then 𝑛 is even.
Rephrased: If 𝑛3 + 5 is odd, then 𝑛 is even
Proof. Assume p is true and q is false (Assume 𝑝. Assume ¬𝑞)
Assume that 𝑛3 + 5 is odd, and 𝑛 is odd.
By the definition of odd numbers 𝑛 = 2𝑘 + 1 for some
integer k.
𝑛3 + 5 = 2𝑘 + 1 3 + 5 = 8𝑘 3 + 12𝑘 2 + 6𝑘 + 6
= 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3)
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Proof by contradiction example
Theorem. If 𝑛 is an integer and 𝑛3 + 5 is odd, then 𝑛 is even.
Rephrased: If 𝑛3 + 5 is odd, then 𝑛 is even
Proof. Assume p is true and q is false (Assume 𝑝. Assume ¬𝑞)
Assume that 𝑛3 + 5 is odd, and 𝑛 is odd.
By the definition of odd numbers 𝑛 = 2𝑘 + 1 for some integer
k.
𝑛3 + 5 = 2𝑘 + 1 3 + 5 = 8𝑘 3 + 12𝑘 2 + 6𝑘 + 6
= 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3)
As 2(4𝑘 3 + 6𝑘 2 + 3𝑘 + 3) is 2 times an integer, it must be
even.
Contradiction!
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End
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