Methods of Proof - Discrete Mathematics and its Applications - Lecture Slides, Slides of Discrete Mathematics

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Lecture 10

1.5 Methods of Proof

1.5 Some Fallacious Proofs

What’s wrong with this?

2

2 2 2

, positive integers

a b

a b

a ab

a b ab b

a b a b b a b

a b b

b b

In this class we will learn the art of proving theorems.

Some names:

1. Theorem, Proposition, Claim, Fact, Result: statement that can be proved.
2. axioms, postulates: the basic assumptions on which the proof us based.
3. lemma: intermediate result to be proved on your way to proof a theorem.
4. corollary: Result that is directly follows from a theorem you just proved.
5. Conjecture: A Result you think is true, but cannot prove.

We use rules of inference to prove theorems. By using them wrong, we create fallacious proofs.

1.5 Rules of Inference

(modus ponens – law of detachment)

[ p^ (^ p^ q^ )] q

p

p q

q

p p q p p q p p q q q

T T T T T

T F F T T F

F T F T T F

F F F T T F

always true: it’s a tautology

Conclusion: if the premises p and pq are both true, then q can only be true. However, if the premises do not hold, q can still be true or false. Docsity.com

Examples:

it snows today If it snows today we go skiing Therefore: we go skiing

[ p ( p q )] q p p q q

∧ → →

→ ∴

If it rains we do not have a barbeque today If we don’t have a barbeque today, we’ll have one tomorrow Therefore: If it rains today, we’ll have a BBQ tomorrow.

[ (^ p^ q^ )^ (^ q^ r^ )^ ] (^ p^ r) p q q r p r

→ ∧ → → → → → ∴ →

1.5 Valid arguments.

All inference rules were of the form: premise 1 is true, premise 2 is true, therefore conclusion is true.

In general this looks like: ( p 1 ∧ p 2 ... ∧ pn ) → q

For an argument to be true all the premises must be true.

Example: if n>1 then n^2 > 1 (True)

We cannot conclude (½)^2 > 1 because the premise is not true.

Fallacies (revisited)

If you do every problem in this book then you’ll learn discrete math. Joe learned discrete math, therefore he did every problem in this book....

p q

p

q

p q

q

p

correct wrong

p = you do all problems in the book. q = do learned discrete math.

fallacy of affirming conclusion

1.5 Inference for Quantified Statements

( ) ( )

xP x P c for arbitrary c

∀

∴

( )

( )

P c for arbitrary c

∴∀ x P x

( ) ( )

xP x P c for some element c

∃ ∴

( )

( )

P c for some element c

∴∃ xP x

universal instantiation (^) universal generalization

existential instantiation existential generalization

Some more examples

Example 10 p.74. All movies produced by John Sayles are wonderful John S. produced a movie about coal-miners Therefore: there a wonderful movie about coalminers.

s(x) = x is a movie by John Sayles c(x) = x is a movie about coalminers w(x) = x is a wonderful movie.

x s x w x

s z w z arbitrary z

s x c x s y c y for some y s y for some y c y for some y

( ) ( ) ( ) ( ( ) ( ))

w y for some y w y c y for some y x w x c x

∧ ∃ ∧

Examples

If it does not rain or it is not foggy then the sailing race will go on and the lifesaving demonstration will go on.

If the sailing is held, then the trophy will be awarded and the trophy was not awarded imply “it rained”.

r = it rains f = it is foggy s = sailing race goes on l = lifesaving demonstration goes on. t = trophy awarded.

( r f ) ( s l)

s t

t

r

t s t s

¬ → ¬

r f s l

s l r f

s l r f

the want the conclusion “r” to be on the right side of the arrow

s l

r f

r

combine premises as much as you can Docsity.com

Strategies for proving theorems

Direct proof of implication pq Assume p = true and use rules of inference to prove that q is true.

Indirect Proof of implication:: Assume q is not true, use rules of inference to prove that p is not true. (NOT q)  (NOT p)

Proof by contradiction: Assume p is not true and use the rules of inference to prove a contradiction. (NOT p)  False

Direct/Indirect Proofs

Proof the following theorem: If n is an odd integer, then n^2 is an odd integer.

assume p (n is an odd integer). n = 2k+1 for some integer k, Then: n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2 ( 2k^2 + 2k) + 1 = 2m+1, m integer 

Proof that if 3n+2 is odd, then n = odd.

Assume (NOT q) : n = even. Then n = 2k, 3n+2 = 6k + 2 = 2(3k+1) = 2m. Thus, 3n+2 is even. We have proved (NOT q)  (NOT p) which is equivalent to pq 