Lecture Notes, 22 Oct 2010
Last Update: 22 October 2010
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§§1.7: Proof Strategies
- Proof by Cases:
- To prove a proposition of the form:
try to show A1 → B
try to show An → B
- i.e., for each i from 1..n do:
- 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
- This is a bit like a case-statement in a programming
- Justification of this strategy:
- [(A1 ∨ … ∨
An) → B] ≡
[(A1 → B)
∧ … ∧ (An → B)]
- Note the switch from "∨" to "∧"
- This logical equivalence cannot be proved by us yet:
- We'll need "mathematical induction"
(the last Peano Axiom)
to do that.
- But note that:
[(A1 ∨ A2) →
B] ≡ [(A1 → B)
∧ (A2 → B)]
You can prove this semantically with a quick t-table to convince yourself of this
or you can prove it syntactically (i.e., formally).
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 n2 ≥ n
- Note that "n is an integer" is logically equivalent
(n < 0) ∨ (n = 0) ∨ (n > 0),
So, to show (n is an integer) →
(n2 ≥ n),
we only have to show:
[(n < 0) ∨ (n = 0) ∨ (n > 0)]
→ (n2 ≥ n)
and this has the form of a proof by cases.
- So, showing that complicated proposition reduces to
showing 3 simpler ones:
- (n < 0) → (n2 ≥ n),
- (n = 0) → (n2 ≥ n),
- (n > 0) → (n2 ≥ n),
- Existence Proofs:
- To prove a proposition of the form:
try one of these kinds of proofs:
- Constructive Proof:
Find c such that P(c).
Then use EG.
- Non-Constructive Proof:
e.g.: Use proof by contradiction:
Suppose ¬∃xP(x), & try to derive a
- 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
but we just don't know which one it is (see IIB):
- e.g.: Thm:
i.e., there are 2 irrational numbers such that when one is
raised to the power of the other, the result is rational (!!)
- Strategy: Find x, y not in Q such that
xy is in Q
- Try x = y = √2.
- Note that √2 is not in Q; i.e.,
- Consider (√2)√2:
For convenience in these notes, call it N.
- Either Q(N) or ¬Q(N).
- Suppose Q(N). Then we're done! We've found our x
Namely: x = y = √2.
- Suppose ¬Q(N).
Then consider N√2, i.e.,
Note that ¬Q(N) (by hypothesis) and ¬Q(√2) (by
Could it be possible that N√2 is rational?
N√2 = (√2)[√2 *
by laws of exponents
and, indeed, Q(2), so we're done!
= (√2)2, by arithmetic
Namely: x = N, and y = √2.
- 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!
- 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.
By the way, although our non-constructive proof doesn't show which one is rational,
the answer is given
- There are lots of other strategies and variations;
we'll introduce them as we need them.
For a summary, see
- Digression: Can any old proposition (or its
negation) be proved?
- There are propositions whose tval we don't know yet
- Up till a few years ago,
Fermat's Last Theorem
not been proved:
- "xn + yn =
zn has no solutions for integers
x, y, z and integer n>2"
but, fairly recently, it was proved.
- "All positive even integers are the sum of 2 primes"
∀x[(Z(x) ∧ (x>0)
∧ Even(x)) →
∧ Prime(z) ∧ (x=y+z)]]
- e.g., 28 = 5 + 23
- This has not (yet) been proved or disproved!
- There are propositions whose tval we know but cannot
- Here's a highly simplified version of it:
- "This proposition is unprovable" is a true proposition
that is unprovable!
And here's an even-more-highly-simplified version of its proof:
- Suppose the quoted proposition is false.
- Then it is false that it is unprovable.
- ∴ it is provable.
- But all provable propositions are true.
- ∴ It is true that it is unprovable.
- ∴ It is both true and unprovable.
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] :-)
Text copyright © 2010 by William J. Rapaport
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