Advanced Discrete Mathematics Jim Skon

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Proofs

• Advanced Discrete MathematicsJim Skon

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Proofs

• Definition A theorem is a valid logical
  assertion whichcan be proved using
• other theorems
• axioms (statements which are given to be true)
  and
• rules of inference (logical rules which allow the
  deduction of conclusions from premises).

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Proofs

• A lemma (not a lemon) is a 'pre-theorem' or a
  result which is needed to prove a theorem.
• A corollary is a 'post-theorem' or a result which
  followsdirectly from a theorem.

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Valid reasoning in proofs

• A mathematical proof is a sequence of statements,
  such that each statement
• 1. is an assumption, or
• 2. is a proposition already proved, or
• 3. Follow logically from one or more previous
  statements in proof.

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Valid Inference

• Consider H1 Ù H2 Ù..... ÙHn C
• where
• Hi are called the hypothesesand
• C is the conclusion.

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Valid Inference Argument

• Argument - consists of a collection of
  statements, called premises of the argument,
  followed by a conclusion statement.
• A1
• A2
• An
• ? A
• where \ means 'therefore' or 'it follows that.'

Premises
Conclusion
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Valid reasoning in proofs

• Example modus ponens
• P Ù (P Q) Þ Q modus ponensP The car is
  running
• Q The car has gas.
• If we know that the car is running (P), we can
  prove that (Q) it has gas.

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Rules of Inference

• Rules of Inference - used in proofs, or
  arguments, to move from what is known to what we
  want to prove.
• modus ponens is a valid rule of inference.

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Valid Argument

• An argument is said to be valid if whenever all
  the premises are true, the conclusion is also
  true.
• If the premises are true, but the conclusion
  false, the argument is said to be invalid.

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Rules of Inference

• Other famous rules of inference P\P Ú Q
  Addition____________________________________P Ù
  Q\PSimplification

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Rules of Inference

• ØQP Q\ØPModus Tollens________________________
  ____________ P QQ R\P RHypothetical
  syllogism

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Rules of Inference

• P ÚQØP\QDisjunctive syllogism
  ____________________________________ PQ\P
  ÙQConjunction

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Rules of Inference

• (P Q) Ù (R S)P Ú R\QÚ SConstructive
  dilemma

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Proofs

• Three Techniques
• Show true using logical inference
• Assume the hypotheses are true
• Use the rules of inference and logical
  equivalences to determine that the conclusion is
  true.
• Show true by showing that no way exists to make
  all premises true but conclusion false
• Show false by finding a way to make premises true
  but conclusion false.

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Formal Proofs

• To prove an argument is valid or the conclusion
  followslogically from the hypotheses
• Assume the hypotheses are true
• Use the rules of inference and logical
  equivalences to determine that the conclusion is
  true.

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Proof Example

• Consider the following logical argument
• If horses fly or cows eat artichokes, then the
  mosquito is the national bird. If the mosquito is
  the national bird then peanut butter takes good
  on hot dogs. But peanut butter tastes terrible on
  hot dogs. Therefore, cows don't eat artichokes.

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Proof Example

• F Horses fly
• A Cows eat artichokes
• M The mosquito is the national bird
• P Peanut butter tastes good on hot dogs

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Proof Example

• Represent the formal argument using the variables
• 1.(FÚ A) M
• 2.M P
• 3.ØP
• \ØA

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Proof Example

• Use the hypotheses 1., 2., and 3. and the above
  rules of inference and any logical equivalencies
  to construct the proof.

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Proof Example

• Assertion Reasons
• 1.(FÚ A) M Hypothesis 1.
• 2.M P Hypothesis 2.
• 3.(F Ú A) P steps 1 and 2 and hypothetical
  syll.
• 4.ØP Hypothesis 3.
• 5.Ø(F Ú A) steps 3 and 4 and modus tollens
• 6.ØF Ù ØA step 5 and DeMorgan
• 7.ØA Ù ØF step 6 and commutativity of 'and'
• 8.ØA step 7 and simplification
• Q. E. D.

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Proof Example

• Consider
• p ? (q ? r)
• q
• ?p ? r

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Proof by contradiction

• Consider
• r ? s
• p ??s
• r ? q
• ?p ? q

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Rules of Inference for Quantifiers

• "xP(x)
• \P(c) Universal Instantiation (UI)
• _______________________________
• P(x)
• \"xP(x) Universal Generalization (UG)

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Rules of Inference for Quantifiers

• P(c)
• \xP(x) Existential Generalization (EG)
• _______________________________
• xP( x)
• \P(c) Existential Instantiation (EI)

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Example

• Every man has two legs. John Smith is a man.
• Therefore, John Smith has two legs.

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Example

• Define the predicatesM(x) x is a manL(x) x
  has two legsJ John Smith, a member of the
  universe
• The argument becomes1."xM(x) L(x)2.M( J
  )\L( J)

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Example

• The proof is1."xM(x) L(x) Hypothesis 12.M( J
  ) L(J ) step 1 and UI3.M(J) Hypothesis 24. L(
  J) steps 2 and 3 and modus ponensQ. E. D.

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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

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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 formP QQ or (P Q) Ù
  Q P \P
• which is not a tautology and therefore not a rule
  of inference!

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The Fallacy of Denying the Antecedent (or the
hypothesis)

• If the butler is nervous, he did it.The butler
  is really mellow.Therefore, the butler didn't do
  it.
• This argument has the formP QØP or (P Q) Ù
  ØP ØQ \ØQwhich is also not a tautology and
  hence not a rule of inference.

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Begging the question or circular reasoning

• This occurs when we use the truth of statement
  being proved (or something equivalent) in the
  proof itself.
• ExampleConjecture if x2 is even then x is
  even.Proof If x2 is even then x2 2k for some
  k. Then x 2l for some l. Hence, x must be even.

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Proof Example

• If the law is sufficient, then Christ died in
  vain
• The law is sufficient
• Therefore Christ died in vain.

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Example

• Babies are illogical
• Nobody is despised who can manage a crocodile
• Illogical persons are despised
• Therefore, babies cannot manage crocodiles

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Methods of Proof

• We wish to establish the truth of the 'theorem
• P Q
• P may be a conjunction of other hypotheses.
• P Q is a conjecture until a proof is produced.

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Types of proof

• Vacuous Proof of P Q
• The truth value of P Q is true if P is false.
  If P can be shown false, then P Q holds.
• Thus prove P Q by showing P is false.
• Example
• If I am both rich and poor then hurricane Fran
  was a mild breeze.
• This is of the form (P Ù ØP) Q
• and the hypotheses form a contradiction.

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Trivial Proof of P Q

• If we know Q is true then P Q is true!
• Example
• If it's raining today then the empty set is a
  subset of every set.
• The assertion is trivially true independent of
  the truth of P.

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Direct Proof of P Q

• Prove Q, using P as an assumption.
• Thus prove P Q by showing Q is true whenever P
  is true.
• assume the hypotheses are true
• use the rules of inference, axioms and any
  logical equivalences to establish the truth of
  the conclusion.
• Example the Cows dont eat artichokes proof above

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Direct Proof of P Q

• Theorem If 6x 9y 101, then x or y is not an
  integer.
• Proof (Direct)
• Assume 6x 9y 101 is true.
• Then from the rules of algebra 3(2x 3y) 101.
• But 101/3 is not an integer so it must be the
  case that one
• of 2x or 3y is not an integer (maybe both).
• Therefore, one of x or y must not be an integer.
• Q.E.D.

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Indirect Proof of P Q

• Prove the contrapositive, e.g. ØQ ØP is true,
  using a direct proof methods.

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Indirect Proof of P Q

• Example
• A perfect number is one which is the sum of all
  its divisors except itself. For example, 6 is
  perfect since 1 2 3 6. So is 28.
• Theorem A perfect number is not a prime.
• Proof (Indirect). We assume the number p is a
  prime and show it is not perfect.
• But the only divisors of a prime are 1 and
  itself.
• Hence the sum of the divisors less than p is 1
  which is not equal to p.
• Hence p cannot be perfect. Q. E. D..

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Proof by contradiction(reductio ad absurdum)

• Assume the negation of the proposition is true,
  then derive a contradiction.
• Thus to prove of P Q, assume P Ù ØQ is true,
  then derive a contradiction.

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Proof by contradiction

• Theorem ?2 is irrational.
• Proof Let P be the proposition ?2 is
  irrational
• Assume ?P or ?2 is rational
• Then ?2 a/b, where a and b are integers and
  have no common factors (lowest terms).
• Then (?2)2 (a/b)2 is 2 a2/b2.
• Thus 2b2 a2. Thus a2 is even, implying a is
  even. Since a is even, a 2c for some integer
  c.
• Thus 2b2 4c2, so b2 2c2. Hence b is even.
• Contradiction! A and b are both even, so
  divisible by 2!

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Proof by cases of P Q

• To prove P Q, find a set of propositions P1,
  P2, ..., Pn, n?2, in which at least one Pj must
  be true for P to be true. P P1? P2 ?
  ... ? Pn
• Then prove the n propositions P1 Q, P2
  Q, ..., Pn Q.

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Proof by cases of P Q

• Let Ä be the operation 'max' on the set of
  integers
• if a ³ b then aÄb maxa, b a bÄa.
• Theorem The operation Ä is associative.
• For all a, b, c
• (aÄb)Äc a Ä(bÄc).

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Proof by casesThe operation Ä is associative

• Let a, b, c be arbitrary integers.
• Then one of the following 6 cases must hold (are
  exhaustive)
• 1. a ³ b ³ c
• 2. a ³ c ³ b
• 3. b ³ a ³ c
• 4. b ³ c ³ a
• 5. c ³ a ³ b
• 6. c ³ b ³ a

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Proof by casesThe operation Ä is associative

• Case 1 aÄb a, aÄc a, and bÄc b.
• Hence
• (aÄb)Äc a aÄ(bÄc).
• Therefore the equality holds for the first case.
• The proofs of the remaining cases are similar.
• Q. E. D.

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Vacuous Proof

• Consider the proposition
• If you your grandfather dies as a baby then you
  will get an A in this class.
• Proof of this statement
• Your grandfather didnt die, thus thus the
  premise must be false. Thus P Q must be true.

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Trivial Proof

• Consider the proposition
• If 3n2 5n -2 ? 2n2 7n - 16 then n n2.
  P(n).
• Proof of P(0)
• 0 02, thus P(0) is trivially true. QED.

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Direct Proof

• Consider The sum of two even numbers is even.
• Restate as
• "x"y (x is even and y is even) x y is even
• Proof
• 1. Remember x is even ax 2a (definition)
• 2. Assume x is even and y is even (assume
  hypothesis)
• 3. x y 2a 2b (from 1 and 2)
• 4. 2a 2b 2(ab)
• 5. By 1, 2(ab) is even - QED.

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Direct Proof

• Consider Every multiple of 6 is also a multiple
  of 3.
• Rewrite "x zy(6x y 3z y)
• Proof
• 1. Assume 6x y (hypothesis)
• 2. 6x y can be rewritten as 3 2x y
• 3. Let z 2x, then 3z y holds. QED.

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Indirect Proofs

• Prove the contrapositive, e.g. Prove that
• ØQ ØP
• is true

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Indirect Proofs

• Prove If x2 is even, then x is even.
• Rewrite "x (EVEN(x2) EVEN(x))

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Indirect Proofs

• Prove If x2 is even, then x is even
• 1. "x (ODD(x) ODD(x2)) (contrapositive)
• 2. Assume 1 ODD(n) true for some n (hypothesis)
• 3. x is odd ax 2a 1 (definition)
• 4. n 2a 1 for some a (2 3)
• 5. n2 (2a 1)2 (substitution)
• 6. (2a 1)2 (2a 1)(2a 1)
• 4a 2 4a 1
• 2 (2a2 2a) 1
• 7. 2 (2a2 2a) 1 is odd (3 6) QED

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Proof by contradiction

• To prove of P Q, assume Ø(P Q), derive a
  contradiction.
• Recall that P Q ? ØP ? Q
• Then Ø(P Q) ? Ø(ØP ? Q) ?
  P Ù ØQ (Demorgans)
• Thus to prove P Q we assume P Ù ØQ and show a
  contradiction.

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Proof by contradiction

• Consider Theorem There is no largest prime
  number.
• This can be stated as
• "If x is a prime number, then there exists
  another prime y which is greater"
• Formally "x y (PRIME(x) Ù PRIME(y) x lt
  y)

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Proof by contradiction

• There is no largest prime number
• Assume largest prime number does exist. Call
  this number p.
• Restate implication as p is prime, and there
  does not exist a prime which is greater.
• 1. Form a product r 2 3 5 ... p)
• (e.g. r is the product of all primes)
• 2. If we divide r1 by any prime, it will have
  remainder 1
• 3. r1 is prime, since any number not divisible
  by any prime which is less must be prime.
• 4. but r1 gt p , which contradicts that p is the
  greatest prime number. QED.

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Proof by cases

• To prove P Q, find a set of propositions P1,
  P2, ..., Pn, n?2, in which at least one Pj must
  be true for P to be true. P P1? P2 ?
  ... ? PnThen prove the n propositions
  P1 Q, P2 Q, ..., Pn Q.
• ThusP(P1ÚP2Ú...ÚPn) and (P1Q)Ù(P2Q)Ù...Ù(PnQ
  )Þ(PQ)

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Proof by cases

• Consider For every nonzero integer x ,x2 gt 0.
• LetP "x is a nonzero integerQ x2 gt 0
• We want to prove P Q

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Proof by cases

• If P "x is a nonzero integer Q x2 gt 0
• Prove P Q
• P can be broken up into two cases
• P1 x gt 0
• P2 x lt 0
• Note that P (P1 Ú P2).

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Proof by cases

• For every nonzero integer x ,x2 gt 0.
• Prove each case -
• Prove P1 Q
• If x gt 0, then x2 gt 0, since the product of
  two positive numbers is always positive.
• Prove P2 QIf x lt 0, then x2 gt 0, since the
  product of two negative numbers is always
  positive. QED.

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Existence Proofs

• We wish to establish the truth of xP( x).
• Constructive existence proof
• Establish P(c) is true for some c in the
  universe.
• Then xP( x) is true by Existential
  Generalization (EG).

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Constructive Existence Proofs

• Theorem There exists an integer solution to the
  equation
• x2 y2 z2 .
• Proof Choose x 3, y 4, z 5.

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Constructive Existence Proofs

• Theorem There exists a bijection from A 0,1
  to B 0, 2.
• Proof We build two injections and conclude there
  must be a bijection without ever exhibiting the
  bijection.
• Let f be the identity map from A to B.
• Then f is an injection (and we conclude that A
  B ).
• Define the function g from B to A as g(x) x/4.
• Then g is an injection.
• Therefore, B A .
• We now apply a previous theorem which states that
• if A B and B A then A
  B .
• Hence, there must be a bijection from A to B.
• (Note that we could have chosen g(x) x/2 and
  obtained a
• bijection directly).
• Q. E. D.

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Nonconstructive existence proof

• Assume no c exists which makes P(c) true and
  derive a contradiction.
• Example
• Theorem There exists an irrational number.
• Proof
• Assume there doesnt exist an irrational number.
• Then all numbers must be rational.
• Then the set of all numbers must be countable.
• Then the real numbers in the interval 0, 1 is a
  countable set.
• But we have already shown this set is not
  countable.
• Hence, we have a contradiction (The set 0,1 is
  countable
• and not countable).
• Therefore, there must exist an irrational number.
• Q. E. D.
• Note we have not produced such a number!

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" Disproof by Counterexample

• Recall that xØP(x) Ø"xP(x ).
• To establish that Ø"xP(x ) is true (or "xP(x) is
  false) construct a c such that ØP(c) is true or
  P(c) is false.
• In this case c is called a counterexample to the
  assertion "xP(x)

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Nonexistence Proofs

• We wish to establish the truth of ØxP( x)
  (which is equivalent to "xØP(x) ).
• Use a proof by contradiction by assuming there is
  a c which makes P(c) true.

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The (infamous) Halting Problem

• We wish to establish the nonexistence of a
  universal debugging program.Theorem There does
  not exist a program which will always determine
  if an arbitrary program P halts.We say the
  Halting Problem is undecidable.

Yes (Halts)
P1
UDP
P2
No (Infinite Loop)
P3
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Halting Problem

• Proof Suppose there is such a program called
  HALT which will determine if any input-free
  program P halts.
• HALT(P) prints 'yes' and halts if P halts,
• otherwise,
• HALT(P) prints 'no' and halts.
• We now construct another procedure as follows
• procedure ABSURD
• if HALT(ABSURD) 'yes' then
• while true do print 'ha'
• (Note that ABSURD is input-free.)

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Halting Problem

• If ABSURD halts then we execute the loop which
  prints unending gales of laughter and thus the
  procedure does not halt.
• If ABSURD does not halt then we will exit the
  program and halt.
• Hence, ABSURD halts if it doesn't
• and doesn't halt if it does
• which is an obvious contradiction.
• Hence such a program does not exist.
• Q. E. D.

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Universally Quantified Assertions

• We wish to establish the truth of
• "xP(x)
• We assume that x is an arbitrary member of the
  universe and show P(x) must be true. Using UG it
  follows that "xP(x) .

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Universally Quantified Assertions

• Example
• Theorem For the universe of integers, x is even
  iff x2 is even.
• Proof The quantified assertion is
• "xx is even x2 is even

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Universally Quantified Assertions

• Proof
• We assume x is arbitrary.
• Recall that P Q is equivalent to (P Q) Ù (Q
  P).
• Case 1. We show if x is even then x2 is even
  using a direct
• proof (the only if part or necessity).
• If x is even then x 2k for some integer k.
• Hence, x2 4k2 2(2k2 ) which is even since it
  is an
• integer which is divisible by 2.
• This completes the proof of case 1.

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Universally Quantified Assertions

• Case 2. We show that if x 2 is even then x must
  be even
• (the if part or sufficiency) .
• We use an indirect proof
• Assume x is not even and show x2 is not even.
• If x is not even then it must be odd.
• So, x 2k 1 for some k.
• Then
• x2 (2k 1)2 4k2 4k 1 2(2k2 2k) 1
• which is odd and hence not even.
• This completes the proof of the second case.
• Therefore we have shown x is even iff x2 is even.
• Since x was arbitrary, the result follows by UG.
• Q.E.D.

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