Proof by exhaustion

Proving a statement by splitting it into a finite, complete set of cases and checking every one — either a short list of values, or all integers split by parity or remainder. Why the cases must leave no gaps, and why exhaustion only works when the cases really are finite.

Pure · Proof

Proof by exhaustion

Why it works

Some statements break into a finite number of cases. If you check every case and each one works, the statement is proved — that is proof by exhaustion. It comes in two shapes:

A short list of values. "For $n \in \mathbb{N}$ , $n \le 4$ , …" gives only the

values

1, 2, 3, 4

. Test each.

Splitting all integers into classes. Every integer is either even or odd;

every integer leaves remainder

0

1

2

on division by

3

(so is

3k

3k + 1

3k + 2

). These classes are infinite in size but finite in number, and together they cover everything — so proving the claim in each class proves it for all integers.

The two non-negotiables:

The cases must be exhaustive — together they must cover every possibility, with

no gaps. Miss a case and you have proved nothing.

Exhaustion needs genuinely finite cases. You cannot "exhaust" an infinite list

by checking the first few values — that is just checking examples, which proves nothing. (Use deduction for an infinite claim with no natural case-split.)

As always, end with a conclusion that the statement holds in all cases.