Discrete Structures

Lecture Notes, 8 Oct 2010

 Last Update: 8 October 2010 Note: or material is highlighted

§§1.6: Proof Strategies

1. Top-Level Proof Strategy:

1. Look at the logical structure
(i.e., at the principal connective or quantifier)
of the proposition to be proved:

• Is it ∀, ∃, →, ¬, etc.?

2. Then use a lower-level strategy for proving propositions of that form.

2. A General Strategy Applicable at Any Time:

• Replace predicates by their definitions whenever that's useful

• i.e., to prove P(x), where P(x) =def D(x),

• e.g., to show Even(n), where Even(x) =def ∃j[x=2j],
try to show ∃j[n=2j]

• That's a search statement, so:
• try to find j such that n=2j

• e.g., to show Odd(n), where Odd(x) =def ∃j[x=2j+1],
try to show ∃j[n=2j]

• That's a search statement, so:
• try to find j such that n=2j+1

3. Direct Proof:

• To show (AB):
Assume/suppose/make believe that A is the case
and try to show B

• Can combine strategies:

• To show (P(n) → B),
where P(x) =def D(x),
suppose D(n) & try to show B

4. To show ∀x[P(x) → Q(x)]:

• Choose an arbitrary object c in the domain,
• try to show (P(c) → Q(c))

• How? By using the Direct Proof strategy above!

• and then apply UG.

5. An Example:

1. Prove: If n is odd, then n+1 is even.
Hidden assumption: n is an integer
Logical form (FOL transation):
(*)   (∀ integer n)[Odd(n) → Even(n+1)]

Strategy & Proof:

• Proofs are stories;
you don't just want to say what happens,
but also how & why

Let c be an arbitrary integer.

• i.e., instead of trying to show that every odd integer is such that its successor is even,
just show it for one "typical" or "arbitrary" integer.

Show Odd(c) → Even(c+1) (and then use UG to show (*))

Suppose Odd(c) & show Even(c+1):

• i.e., instead of showing that the successor of odd c is even,
suppose an extra fact, namely, that c is odd,
& just show that c+1 is even

i.e., show ∃j[c+1=2j] (replace "Even" by its definition)
But Odd(c)
∴ ∃k[c=2k+1] (by def of Odd)

• (using "k" to avoid confusion with "j")

Call the k that works "k1" (by EI)
Now show (2k1+1)+1 is even:

Use algebra: (2k1+1)+1 = 2k1+2 = 2(k1+1)
So, take j = k1+1
i.e., ∃j[c+1=2j], by EG, namely: j = k1+1
c+1 is even,
i.e., Even(c+1).
Now that we know Even(c+1) on the assumption that Odd(c),
we also know Odd(c) → Even(c+1)
∴ we can infer (∀ integer n)[Odd(n) → Even(n+1)], by UG
QED

2. Here is a more formal ("cleaned up") version:

Show: (∀ integer n)[Odd(n) → Even(n+1)]
Let c be an arbitrary integer
Show Odd(c) → Even(c+1):

 1 Odd(c) :  temporary assumption for Direct Proof 2 ∃k[c=2k+1] :  1; def of Odd 3 c=2k1+1 :  2; EI 4 c+1=2k1+1+1 :  3; algebra
…(cue the closing music and a deep voice: "to be continued next time")