Discrete Structures

# Lecture Notes, 22 Oct 2010

 Last Update: 22 October 2010 Note: or material is highlighted

### §§1.7: Proof Strategies

1. Proof by Cases:

1. To prove a proposition of the form:

(A1 ∨ … ∨ An) → B

try to show A1B and … and try to show AnB

1. i.e., for each i from 1..n do:
try to show AiB

2. I.e., to show that, if any of the Ai are true, then B is true,
try to show that each of the Ai imply B.

3. This is a bit like a case-statement in a programming language!

2. Justification of this strategy:

1. [(A1 ∨ … ∨ An) → B] ≡ [(A1B) ∧ … ∧ (AnB)]

2. Note the switch from "∨" to "∧"

3. This logical equivalence cannot be proved by us yet:

1. We'll need "mathematical induction"
(the last Peano Axiom)
to do that.

2. But note that:

[(A1A2) → B] ≡ [(A1B) ∧ (A2B)]

You can prove this semantically with a quick t-table to convince yourself of this
or you can prove it syntactically (i.e., formally).

4. There is sometimes a trick in applying this,
because the antecedent of a theorem might not look as though it's a disjunction:

• e.g.: Thm: If n is an integer, then n2n

1. Note that "n is an integer" is logically equivalent to:

(n < 0) ∨ (n = 0) ∨ (n > 0),

So, to show (n is an integer) → (n2n),
we only have to show:

[(n < 0) ∨ (n = 0) ∨ (n > 0)] → (n2n)

and this has the form of a proof by cases.

2. So, showing that complicated proposition reduces to showing 3 simpler ones:

1. (n < 0) → (n2n), and
2. (n = 0) → (n2n), and
3. (n > 0) → (n2n),

• Case (a):
Suppose n < 0, & show n2n:
Because n < 0, we have n2 > 0
(the product of 2 negatives is positive)
But n < 0 implies 0 > n,
so 0 ≥ n, by the Addition rule of inference(!)
so n2 > 0 and 0 ≥ n (by Conjunction)
so n2n (by transitivity of ≥)

• Case (b):
Suppose n = 0, & show n2n:
Because n = 0, we have n2 = 0,
so n2 = n
(because they're both = 0)

• Case (c):
Suppose (n > 0), and show n2n:
Because (n > 0), we have n2 > 0,
but that doesn't seem to get us anywhere :-(
So, try this:
Because (n > 0), we have n ≥ 1
So, n*n ≥ 1*n (multiplying both sides by n)
So, n2n.

Case (c) is a nice example of a tricky proof
where it's more important for you to understand the proof
than it is to worry about how you would have come up with the trick at the end of it.

QED

2. Existence Proofs:

1. To prove a proposition of the form:

xP(x)

try one of these kinds of proofs:

1. Constructive Proof:
Find c such that P(c).
Then use EG.

2. Non-Constructive Proof:
Suppose ¬∃xP(x), & try to derive a contradiction.

• I.e., instead of finding x, show that it has to exist (somewhere) on pain of contradiction.

Or: Show that one of two (or more) candidates for x satisfies P,
but we just don't know which one it is (see IIB):

2. e.g.: Thm: ∃xy[¬Q(x) ∧ ¬Q(y) ∧ Q(xy)]

i.e., there are 2 irrational numbers such that when one is raised to the power of the other, the result is rational (!!)

proof:

1. Strategy: Find x, y not in Q such that xy is in Q
2. Try x = y = √2.
3. Note that √2 is not in Q; i.e., ¬Q(√2) (by earlier proof)
4. Consider (√2)√2: For convenience in these notes, call it N.
5. Either Q(N) or ¬Q(N).

6. Suppose Q(N). Then we're done! We've found our x and y!
Namely: x = y = √2.

7. Suppose ¬Q(N).
Then consider N√2, i.e., [(√2)√2]√2
Note that ¬Q(N) (by hypothesis) and ¬Q(√2) (by earlier proof).
Could it be possible that N√2 is rational?
N√2 = (√2)[√2 * √2], by laws of exponents
= (√2)2, by arithmetic
= 2
and, indeed, Q(2), so we're done!
Namely: x = N, and y = √2.

8. But which is it? Is the answer step 6 or step 7?

WE DON'T KNOW!

But one of them works, and that's all that we were asked to prove!
QED

3. This is an example of a non-constructive proof,
because we didn't really find (or "construct") the answer;
we merely showed that there is an answer.

Not all mathematicians or computer scientists accept non-constructive proofs.

By the way, although our non-constructive proof doesn't show which one is rational, the answer is given here :-)

3. There are lots of other strategies and variations;
we'll introduce them as we need them.

For a summary, see Proof Strategies

4. Digression: Can any old proposition (or its negation) be proved?

NO!

1. There are propositions whose tval we don't know yet

1. Up till a few years ago, Fermat's Last Theorem had not been proved:

• "xn + yn = zn has no solutions for integers x, y, z and integer n>2"

but, fairly recently, it was proved.

2. Goldbach's Conjecture:

• "All positive even integers are the sum of 2 primes"

1. i.e.: ∀x[(Z(x) ∧ (x>0) ∧ Even(x)) → ∃yz[Prime(y) ∧ Prime(z) ∧ (x=y+z)]]

2. e.g., 28 = 5 + 23

3. This has not (yet) been proved or disproved!

2. There are propositions whose tval we know but cannot prove!

• Gödel's Incompleteness Theorem:

1. Here's a highly simplified version of it:

• "This proposition is unprovable" is a true proposition that is unprovable!

2. And here's an even-more-highly-simplified version of its proof:

1. Suppose the quoted proposition is false.
2. Then it is false that it is unprovable.
3. ∴ it is provable.
4. But all provable propositions are true.
5. ∴ It is true that it is unprovable.
6. ∴ It is both true and unprovable.

3. Compare Shakespeare:
There are more things [that are true] in heaven and earth than are dreamt of in your philosophy [i.e., that can be proved by your logic] :-)