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

By Cardinality of Empty Set:
 * $X = \varnothing$

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\}$

By Union of Unordered Tuples
 * $ X = \left\{ {x_1, \dots, x_n}\right\} \cup \left\{ {x_{n+1} }\right\}$

By definition of cardinality:
 * $\left\vert{\left\{ {x_1, \dots, x_n}\right\} }\right\vert = n$

By Set is Subset of Union:
 * $\left\{ {x_1, \dots, x_n}\right\} \subseteq X \subseteq A$

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

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