Discrete Structures

# Lecture Notes, 13 Oct 2010

 Last Update: 13 October 2010 Note: or material is highlighted

### §§1.6: Proof Strategies (cont'd)

1. The Example from Last Time, concluded:

To show any proposition A:
supp. ¬A & try to derive a contradiction;
i.e., try to find B such that (B ∧ ¬B)

2. E.g.: Show √2 is irrational:

• Let Q(x) = "x is rational"

• Remember: Q is the mathematical symbol for the set of rational numbers.

So we need to show: ¬Q(√2)

• Suppose, by way of contradiction, Q(√2)
& try to find B such that (B ∧ ¬B):

• Because we are assuming that Q(√2),
it follows that (∃ int a,b)[(√2 = a/b) ∧ (a/b is in lowest terms)],

• i.e., a,b have no common factors

• Call this B

∴ 2 = a2/b2

∴ 2b2 = a2

∴ Even(a2)

∴ Even(a), by lemma (Rosen, p.85: 16)

• (Hint: If Even(a2), could Odd(a)?)

∴ ∃j[a=2j]; by EI, call it j1

∴ 2b2 = (2j1)2 = 4j12

b2 = 2j12

∴ Even(b2)

∴ Even(b)

But now we have Even(a) ∧ Even(b).

a,b do have a common factor, namely: 2

• i.e., ¬B!

So, we have our contradiction, (B ∧ ¬B)
which means that our assumption for the proof by contradiction, Q(√2), was wrong.

∴ ¬Q(√2). QED

2. More Strategies:

1. (not covered in lecture, but included here for reference)

Another strategy to prove (AB)
(another form of proof by contradiction):

• Suppose A
Suppose ¬B
& try to find C such that (C ∧ ¬C)

Works because:

• previous strategy for (AB) was to suppose A & show B, and
• previous strategy to show B was to suppose ¬B & try to derive a contra.
• this strategy combines them.

2. To prove (AB):

• Show (AB), and
show (BA)

3. To prove (A1A2A3 ↔ … ↔ An):

• could try to show that all possible pairs are equivalent, but it's easier to…

show (A1A2), and
show (A2A3), and
…, and
show (An–1An), and
show (AnA1)

• E.g.: The Church-Turing Thesis says that all models of computation are equivalent to each other:

Turing machines ↔ lambda-calculus ↔ Post production systems ↔ Markov algorithms ↔ register machines, etc.

To prove this, compile each into the next, and the last into the first.

4. To prove ¬∀xP(x):

• Find (i.e., search for; i.e., ∃) a counterexample
i.e., show ∃x¬P(x)

3. There are lots of ways to have invalid proofs!

1. See Rosen, pp. 83–84.

2. More importantly:

1. It's possible for there to be invalid proofs of true conclusions!

• That doesn't make the proof valid!
• And it doesn't prove the conclusion!

• E.g.:

All birds fly.
Tweety the canary flies.
∴ Tweety is a bird.

• The structure of this argument is:

x[P(x) → Q(x)]
Q(c)
∴ P(c)

• This is an invalid argument with all true premises
and a true conclusion
(at least, according to Warner Bros. cartoons!)

• To show it's invalid, consider this argument with:

• the same structure,
• all true premises,
• but a false conclusion:

All dogs are mammals.
Garfield the cat is a mammal.
∴ Garfield is a dog.

2. And it's possible for there to be valid proofs of false conclusions!
(But they're not sound!)

• E.g.:
2+2=5
(2+2=5) → Pope(bertrand-russell)
∴ Pope(bertrand-russell)

proof #1:

This is argument is an instance of the valid argument form modus ponens!

proof #2: Show (2+2=5) → Pope(bertrand-russell) by Direct Proof:

Suppose 2+2=5 & show Pope(b-r):

 1 2+2=5 : assumption 2 (2+2=5) ∨ Pope(b-r) : 1; Addition 3 ¬(2+2=5) : theorem of arithmetic 4 Pope(b-r) : 2,3; DS

• This argument works because of the truth-table for the material conditional,
which, as I said when I introduced it, doesn't exactly match the conditional of ordinary English.

• For more on this, see "Paradoxes of Material Conditional"

• The branch of logic that disallows this is called "relevance logic".