# Induction of Finite Set

## Theorem Scheme

Let $A$ be finite set.

Let $P\left({-}\right)$ be a predicate.

Let $P\left({\varnothing}\right)$

Let

$\forall B \subseteq A, x \in A: \left({ P\left({B}\right) \implies P\left({B \cup \left\{ {x}\right\} }\right)}\right)$

Then

$P\left({A}\right)$

## Proof

We will prove the result by induction on cardinality of argument.

### Base Case

$\forall X \subseteq A: \left({ \left\vert{X}\right\vert = 0 \implies P\left({X}\right)}\right)$

Let $X \subseteq A$ such that

$\left\vert{X}\right\vert = 0$
$X = \varnothing$

Thus by assumption:

$P\left({X}\right)$

### Induction Hypothesis

$\forall X \subseteq A: \left({ \left\vert{X}\right\vert = n \implies P\left({X}\right)}\right)$

### Induction Step

$\forall X \subseteq A: \left({ \left\vert{X}\right\vert = n+1 \implies P\left({X}\right)}\right)$

Let $X \subseteq A$ such that

$\left\vert{X}\right\vert = n+1$

By definition of cardinality:

$X = \left\{ {x_1, \dots, x_n, x_{n+1} }\right\}$
$X = \left\{ {x_1, \dots, x_n}\right\} \cup \left\{ {x_{n+1} }\right\}$

By definition of cardinality:

$\left\vert{\left\{ {x_1, \dots, x_n}\right\} }\right\vert = n$
$\left\{ {x_1, \dots, x_n}\right\} \subseteq X \subseteq A$

Then by Induction Hypothesis:

$P\left({\left\{ {x_1, \dots, x_n}\right\} }\right)$

By definition of subset:

$x_{n+1} \in A$

Thus by assumption:

$P\left({X}\right)$

$\Box$

By Induction Thesis:

$\forall X \subseteq A: \left({ \left\vert{X}\right\vert = \left\vert{A}\right\vert \implies P\left({X}\right)}\right)$

Hence

$P\left({A}\right)$

$\blacksquare$