Methods of Proof

1
Methods of Proof

• Section 1.5

2
Definitions

• A theorem is a valid logical assertion which can
  be proved using
• Axioms statements which are given to be true
• Rules of inference logical rules allowing the
  deduction of conclusions from premises
• A lemma is a pre-theorem or a result which is
  needed to prove a theorem.
• A corollary is a post-theorem or a result which
  follows directly from a theorem.

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Rules of Inference
H1 H2 Hn ?? ? C

• H1, H2, Hn are the hypotheses
• We use conjunction H1 H2 H3

C is the conclusion. ? means therefore or
it follows that
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Some Rules of Inference
Mode that affirms
5
Example Simplification

• pit is below freezing
• qit is raining now
• It is below freezing and raining now.
• Therefore, it is below freezing.

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Example Modus Ponensfrom Latin mode that
affirms

• In other words
• If the hypothesis p is true
• and the hypothesis (p-gtq) is true
• Then I can conclude q

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Example Modus Ponens

• p n is greater than 3
• q n2 is greater than 9
• Assuming that p is true, and p? q is true, then
• if is n greater than 3, it follows that n2 is
  greater than 9.

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More Rules of Inference
mode that denies
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Exercise

• If It rains today, then we will not have a
  barbecue today. If we do not have a barbecue
  today, then we will have a barbecue tomorrow
• Therefore, if it rains today, then we will have a
  barbecue tomorrow.

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Validity of an Argument

• An argument is valid if
• whenever all hypotheses are true, the conclusion
  is also true
• To prove that an argument is valid
• Assume the hypotheses are true
• Use the rules of inference and logical
  equivalences to determine that the conclusion is
  true

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Recap 1.2 Important Equivalences
12
Recap 1.2 Important Equivalences
13
Recap 1.2 Important Equivalences
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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 tastes good
  on hot dogs. But peanut butter tastes terrible on
  hot dogs. Therefore, cows dont eat artichokes.
• Assign propositional variables to each component
  in the argument

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Example

• Assignments
• p Horses fly
• q Cows eat artichokes
• r The mosquito is the national bird
• s Peanut butter tastes good on hot dogs

• Represent the argument using the variables
• (p ? q) ? r
• r ? s
• ?s
• ? ?q Conclusion

Hypotheses
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Example

• Assertion Reasons

1. (p ? q) ? r Hypothesis
2. r ? s Hypothesis
3. (p ? q) ? s Hypothetical syll. on 1. and 2.
4. ?s Hypothesis
5. ?(p ? q) Modus tollens on 3. and 4.
6. ?p ? ?q DeMorgan on 5.
7. ?q ? ?p Commutative on 6.
8. ?q Simplification on 7.
We got our conclusion of cows dont eat
artichokes
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Example

• Show that the following argument is valid
• It is not sunny this afternoon and it is colder
  than yesterday. We will go swimming only if it is
  sunny. If we do not go swimming, then we will
  take a canoe trip. If we take a canoe trip, then
  we will be home by sunset. Therefore, we will be
  home by sunset.

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Rules of Inference for Quantified Statements
?xP(x), then for any C, therefore P(c) is true
we select any element c and P(c) is
true Therefore ?xP(x)
?xP(x) therefore for at least one specific
c, P(c) is true
we select a particular element c and P(c) is
true Therefore, ?xP(x)
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Example

• Everyone in the discrete math class has taken a
  CS course. Marla is a student in the discrete
  class. Therefore, Marla has taken a CS course.

D(x) x is in the discrete math class C(x) x has
taken a CS course
1. ?x(D(x) ? C(x)) Premise 2. D(Marla) ?
C(Marla) Univ. Inst. using 1. 3.
D(Marla) Premise 4. C(Marla) Modus ponens
using 2. and 3.
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Exercise

• A student in this class has not read the book.
  Everyone in this class passed the first exam.
• Therefore, Someone who passed the first exam has
  not read the book

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Fallacies

• Fallacies resemble rules of inference but are
  based on contingencies rather than tautologies.
  They are incorrect inferences.
• Three common fallacies
• Affirming the Consequent
• Denying the Hypothesis
• Circular Reasoning (begging the question)

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Fallacy of Affirming the Consequent

• ((p ? q) ? q) ? p is not a tautology and
  therefore not a rule of inference.
• If you do every problem in this book, then you
  will learn discrete mathematics. You learned
  discrete mathematics. Therefore, you did every
  problem in this book.
• Exercise. Compare with modus ponens and tollens

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Fallacy of Denying the Hypothesis

• ((p ? q) ? ?p) ? ?q is not a tautology and
  therefore not a rule of inference.
• If you do every problem in this book, then you
  will learn discrete mathematics. You did not do
  every problem in this book. Therefore, you did
  not learn discrete mathematics.
• Exercise. Compare with modus ponens and tollens

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Circular Reasoning

• One or more steps of the proof are based upon the
  truth of the statement being proved.
• Also known as begging the question

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Exercises

• 1, 3, 4, 7, 9, 11

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RecapSome Rules of Inference
Mode that affirms
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RecapMore Rules of Inference
mode that denies
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RecapRules of Inference for Quantified
Statements
?xP(x), then for any C, therefore P(c) is true
we select any element c and P(c) is
true Therefore ?xP(x)
?xP(x) therefore for at least one specific
c, P(c) is true
we select a particular element c and P(c) is
true Therefore, ?xP(x)
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Methods of Proof

• Direct proof
• Indirect proof
• Vacuous proof
• Trivial proof
• Proof by contradiction
• Proof by cases
• Existence proof

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Proof Basics

• We want to establish the truth of p ? q
• p may be a conjunction of other hypotheses
• p ? q is a conjecture until a proof is produced

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Direct Proof

• Assume the hypotheses are true
• Use rules of inference and any logical
  equivalences to establish the truth of the
  conclusion
• HOW TO PROVE
• If p is true ,then q has to be true for pgtq to
  be true
• Example The proof we did earlier about cows not
  eating artichokes was an example of a direct
  proof

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Example

• Give a direct proof of the theorem If n is an
  odd integer, then n2 is an odd integer
• (n is odd) ? (n2 is odd)
• Using the following definition
• If n is even, then exist an integer k such that
  n2k, and It is odd, if there exist and integer
  k such that n2k1.

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Example (Cont)

• Assume the hypothesis n is odd true
• n is odd
• Since n is odd, then ?k n2k1
• Now, is the conclusion n2 is odd true?
• n2 (2k1)2 4k2 4k 1
• 2(2k22k)1
• 2 (m) 1, where some integer m2k22k
• Since n2 2(m)1, then n2 is odd is true
• Proof complete

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Indirect Proof

• A direct proof of the contrapositive
• Remember p?q is equivalent to q ? p
• Proof q ? p
• Assume that ?q is true i.e., q is false
• Use rules of inference and logical equivalences
  to show that ?p is true i.e., p is false

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Example

• Give an indirect proof to the theorem if 3n2 is
  odd, then n is odd
• (3n2 is odd) ? (n is odd)
• p 3n2 is odd, p 3n2 is even
• q n is odd, q n is even
• The contrapositive is
• (n is odd) ? (3n2 is odd) , in other words
• (n is even) ? (3n 2 is even)

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Example (Cont)

• Assuming the hypothesis (of the contra positive)
  n is even true
• Then n2k
• Now, is the conclusion (of the contrapositive)
  3n2 is even true?
• 3n2 3(2k)26k2
• 2(3k1)
• 2(m), where m 3k1
• Then 3n2 is even is true
• Proof complete

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Vacuous Proof

• If we know one of the hypotheses in p is false
  then p?q is vacuously true.
• F ? T and F ? F are both true.
• Example
• If I am both rich and poor, then hurricane
  Katrina was a mild breeze.
• The hypotheses (p??p) form a contradiction,
  therefore q follows from the hypotheses
  vacuously.
• Sometimes used to proof theorems using UNIVERSAL
  QUANTIFICATION

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Example

• Show that P(0) is true where P(n) If n gt 1, then
  n2 gt n.
• P(n) (ngt1) ? (n2 gt n)
• P(0) (0gt1) ? ( 02 gt 0)
• Since the hypothesis (0gt1) is false, P(0) is
  automatically true.
• Note that we do not even pay attention to the
  conclusion 02 gt 0

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Trivial Proof

• If we know q is true, then p?q is true
• F ? T and T ? T are both true.
• Example
• If its raining today then the empty set is a
  subset of every set.
• The assertion is trivially true independent of
  the truth value of p.

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Example

• Show that P(0) is true where P(n) If a ? b gt 0,
  then an ? bn.
• P(n) (a ? b gt 0) ? (an ? bn)
• P(0) (a? b gt 0) ? (a0 ? a0), in other words
• P(0) (a? b gt 0) ? (1 ? 1),
• Since the conclusion (1 ? 1) is true, hence P(0)
  is true.
• Note that we do not even pay attention to the
  hypothesis (a? b gt 0)

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Proof by ContradictionReductio ad absurdum
(reduction to the absurd )

• We want to proof that p?q, but what if we can
  proof that p implies a contradiction (i.e., q is
  FALSE no matters what, or an absurd)??
• Mathematical definition of the proof
• Find a contradiction q such that
• ?p?q ? ?p?F ? T
• Consequently, if we show the contradiction, then
  the assumption p is wrong (FALSE), so p must be
  true
• To prove that p is true, you have to show that ?p
  leads to a contradiction i.e., you have to prove
  that ?p?F is true.

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Example

• Prove that v2 is irrational
• P v2 is irrational
• What if p is true, v2 is rational
• Does this lead to a contradiction???
• If v2 is rational, then ?a,b integers such that
  v2 a /b (assuming ra and b have no common
  factors)
• v2 a/ b , then 2 a2/b2, then 2b2a2

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Example cont

• Since a22 (b2), then a2 is even, therefore a is
  even, then
• 2b2 (2c)2, for some integer c
• 2b24c2, so b22c2
• Thus, b is even too.
• If a and b are even, then they have at least one
  common factor (2), so the assumption r is
  contradicted p-gt (rr)
• Therefore, p is false, p is true
• v2 is irrational is true

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Proof by Contradiction (Cont..)

• An indirect proof of an implication p?q can be
  rewritten as a proof by contradiction.
• Assume that both p and ?q are true.
• Then use a direct proof to show that
• ?q ? ?p
• This leads to the contradiction p??p.
• HOMEWORK. Example 22 (pg 67) If 3n2 is odd, then
  n is odd.

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Proof by Cases

• Break the premise of p?q into an equivalent
  disjunction of the form p1?p2???pn
• Then use the equivalence
• (p1?p2???pn) ? q ? (p1?q)?(p2 ?q) ? ? ?(pn?q)
• Each of the implications pi ?q is a case.
• You must
• Convince the reader that the cases are inclusive
  (i.e., they exhaust all possibilities)
• Establish all implications

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Proof by Cases (Cont..)

• If n is an integer, then n21 is positive.
• p n is an integer
• q n21 is positive
• p ? (p1?p2?p3)
• where, p1 n 0
• p2 n gt 0
• p3 n lt 0
• Now prove that ((p1?q)?(p2 ?q)?(p3 ?q)) is true
  i.e., all the cases (p1?q), (p2 ?q), and (p3 ?q)
  are true.

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Proving an equivalence

• To prove that p?q, you need to show that p?q is a
  tautology.
• You can do that by showing that p?q and q?p are
  both true since,
• p?q ? (p?q)?(q?p)
• Example The integer n is odd if and only if n2
  is odd.

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Existence Proof

• The proof of ?xP(x) is called an existence proof.
• Constructive
• Non-constructive
• Constructive existence proof
• Find an element c in the universe of discourse
  such that P(c) is true
• Non-constructive existence proof
• Do not find c, rather, somehow prove ?xP(x) is
  true
• Generally, by contradiction
• Assume no c exists that makes P(c) true
• Derive a contradiction