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cs2102: Discrete Mathematics
Class 2: Proof Methods
David Evans
University of Virginia
If you have any questions about
syllabus, course organization,
etc. ask them in #inclass now
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Plan
Review proof from end of last class
How rigorous should proofs be?
Aside: “Electronic Signatures”
Proof Methods:
– “If and only if”
What makes a “good” proof?
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If the product of x and y is even, at least one of x or y must be even.
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If the product of x and y is even, at least one of x or y must be even.
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If the product of x and y is even, at least one of x or y must be even.
P = “the product of x and y is even”
Q = “at least one of x or y must be even”
What definitions do we need?
Goal: prove P implies Q
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If the product of x and y is even, at least one of x or y must be even.
P = “the product of x and y is even”
Q = “at least one of x or y must be even”
Goal: prove P implies Q
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If the product of x and y is even, at least one of x or y must be even.
P = “the product of x and y is even”
Q = “at least one of x or y must be even”
Goal: prove P implies Q
Definition: even. An integer, z, is even if there
exists an integer k such that z = 2k.
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If the product of x and y is even, at least one of x or y must be even.
P = “the product of x and y is even”
Q = “at least one of x or y must be even”
Goal: prove P implies Q
Definition: even. An integer, z, is even if there
exists an integer k such that z = 2k.
Definition: odd. An integer, z, is odd if there exists
an integer k such that z = 2k + 1.
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Definition: even. An integer, z, is even if and only
if there exists an integer k such that z = 2k.
Definition: odd. An integer, z, is odd if and only if
there exists an integer k such that z = 2k + 1.
Odd-Even Lemma: If an integer is not even, it is odd.
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Definition: even. An integer, z, is even if and only
if there exists an integer k such that z = 2k.
Definition: odd. An integer, z, is odd if and only if
there exists an integer k such that z = 2k + 1.
Odd-Even Lemma: If an integer is not even, it is odd.
Prove an implication: assume P, show chain that leads to Q.
Assume z is an integer that is not even.
Substitute definition:
NOT(exists integer k such that z = 2k)
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Lemma: If an integer is not even, it is odd.
Prove an implication: assume P, show chain that leads to Q.
Assume z is an integer that is not even.
Substitute definition:
NOT(exists integer k such that z = 2k)
Meaning of NOT:
no integer k exists such that z = 2k
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Lemma: If an integer is not even, it is odd.
Prove an implication: assume P, show chain that leads to Q.
Assume z is an integer that is not even.
Substitute definition:
NOT(exists integer k such that z = 2k)
Meaning of NOT:
no integer k exists such that z = 2k
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“Electronic Signatures” Digression
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Ritualistic Signatures
• Intended to make signer
think seriously and impact
signer’s behavior
• Varying impact (?):
– Quill on parchment
– Pen on paper
– Scribbling on tablet
– Typing
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Ritualistic Signatures
• Intended to make signer
think seriously and impact
signer’s behavior
• Varying impact (?):
– Quill on parchment
– Pen on paper
– Scribbling on tablet
– Typing
Purposeful Signatures
• Verifiable
• Non-repudiable
• Bound to content
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No content
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Purposeful Signatures
Verifiable
Non-repudiable
Bound to content
Real electronic signatures:
One-way hard problem
Easy to raise to powers,
hard to find discrete logs
Signature combines message
and private key
Can be verified by obtaining
public key from trusted source
and checking signature is valid
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Signing Pledges
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If the product of x and y is even, at least one of x or y must be even.
P = “the product of x and y is even”
Q = “at least one of x or y must be even”
Goal: prove P implies Q
Definition: even. An integer, z, is even if there
exists an integer k such that z = 2k.
Lemma: If an integer is not even, it is odd.
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
To prove, P implies Q, we use contrapositive inference rule:
NOT(Q) implies NOT(P)
P implies Q
Observe: this is starting backwards!
We are starting the proof from the conclusion we want.
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
To prove, P implies Q, we prove the contrapositive:
NOT(Q) implies NOT(P)
To prove an implication, assume left side, show right:
Assume NOT(at least one of x and y must be even)
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
To prove, P implies Q, we prove the contrapositive:
NOT(Q) implies NOT(P)
To prove an implication, assume left side, show right:
Assume NOT(at least one of x and y must be even)
By the meaning of NOT:
both x and y are not even
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
To prove, P implies Q, we prove the contrapositive:
NOT(Q) implies NOT(P)
To prove an implication, assume left side, show right:
Assume NOT(at least one of x and y must be even)
By the meaning of NOT:
both x and y are not even
By the Odd-Even Lemma:
both x and y are odd
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
To prove, P implies Q, we prove the contrapositive:
NOT(Q) implies NOT(P)
To prove an implication, assume left side, show right:
Assume NOT(at least one of x and y must be even)
By the meaning of NOT:
both x and y are not even
By the Odd-Even Lemma:
both x and y are odd
By the definition of odd:
there exists integers k, m, such that x = 2k + 1 and y = 2m + 1
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
We prove the contrapositive:
Assume NOT(at least one of x and y must be even)
By the meaning of NOT: both x and y are not even.
By the Odd-Even Lemma: both x and y are odd.
By the definition of odd:
there exists integers k, m, such that x = 2k + 1 and y = 2m + 1
By algebra:
xy = (2k + 1)(2m + 1) = 4mk + 2m + 2k + 1
= 2(2mk + m + k) + 1
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
We prove the contrapositive:
Assume NOT(at least one of x and y must be even)
…
By integers closed under multiplication and addition:
there exists some integer r where r = 2mk + m + k
By definition of odd:
So xy = 2r + 1 which means the product of x and y is odd.
By the Odd-Even Lemma:
the product of x and y is not even.
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
We prove the contrapositive:
Assume NOT(at least one of x and y must be even)
…
By integers closed under multiplication and addition:
there exists some integer r where r = 2mk + m + k
By definition of odd:
So xy = 2r + 1 which means the product of x and y is odd.
By the Odd-Even Lemma:
the product of x and y is not even.
Lemma: If an integer is not even, it is odd.
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
We prove the contrapositive:
Assume NOT(at least one of x and y must be even)
…
By the Even-Odd* Lemma:
the product of x and y is not even.
So, NOT(product of x and y is even)
Thus, we have proven the implication:
NOT(at least one of x and y must be even)
implies NOT(product of x and y is even)
As shown in class, this
slide had bugs in it - the
version here is fixed
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P = “the product of x and y is even”
Q = “at least one of x or y must be even”
We prove the contrapositive:
Assume NOT(at least one of x and y must be even)
…
So, NOT(product of x and y is even)
Thus, we have proven the implication:
NOT(at least one of x and y must be even)
implies NOT(product of x and y is even)
By the contrapositive inference rule, this proves:
at least one of x and y must be even
implies product of x and y is even.
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Ay, Harambe!
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Purpose of a Proof
Seeking Truth
Deepening Understanding
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In physics, your solution
should convince a
reasonable person. In
math, you have to
convince a person who's
trying to make trouble.
Frank Wilczek
(2004 Nobel Prize Physics)
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“Correct ” Rigorous Proof of P
Axioms used are clear and accepted
Each step uses a sound inference rule correctly:
– Shows antecedents are satisfied
– Concludes the conclusion
Results in concluding goal proposition: P
Do the proofs we do in cs2102 actually do this?
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“Good” Proofs in cs2102
• Well written and clearly organized:
– Should be obvious what you are proving and how
• Convincing to a skeptical reader
• State assumptions clearly: careful about not
assuming non-obvious things
• Focus on important steps, not gory details
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Whitehead and Russell, Principia Mathematica
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Automated Theorem Proving
Coq
PVS
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Example Automated Proofs
(thm (implies
(and (evenp x) (evenp y))
(evenp (+ x y))))
From https://rosettacode.org/wiki/Proof
ACL2
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Inductive nat : Set :=
| O : nat
| S : nat -> nat.
Inductive even : nat -> Set :=
| even_O : even O
| even_SSn : forall n:nat, even n -> even (S (S n)).
Theorem even_plus_even :
forall n m:nat, even n -> even m -> even (n + m).
Proof.
… (about 10 lines)
Qed.
From https://rosettacode.org/wiki/Proof
Coq
Human-assisted, automatically verified
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Automatic Theorem Provers in
Common Practice?
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Proposition: input program does not have certain types of errors.
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Processor Verification
FDIV 0x4005FB / 0x2FFFFF
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“Miracle on the Hudson” ?
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Charge
• Next week: we’ll start proving non-obvious
things!
• Before Friday (6:29pm): course pledge,
registration survey (read “Habits of Highly
Mathematical People”)
• Read MCS Ch 2, 3; PS1 is due next Friday
• TA’s office hours will start Monday