Induction of Finite Set

Theorem Scheme
Let $A$ be finite set.

Let $\map P -$ be a predicate.

Let $\map P \O$.

Let
 * $\forall B \subseteq A, x \in A: \paren {\map P B \implies \map P {B \cup \set x} }$

Then:
 * $\map P A$

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

Base Case

 * $\forall X \subseteq A: \paren {\size X = 0 \implies \map P X}$

Let $X \subseteq A$ such that:
 * $\size X = 0$

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

Thus by assumption:
 * $\map P X$

Induction Hypothesis

 * $\forall X \subseteq A: \paren {\size X = n \implies \map P X}$

Induction Step

 * $\forall X \subseteq A: \paren {\size X = n + 1 \implies \map P X}$

Let $X \subseteq A$ such that:
 * $\size X = n + 1$

By definition of cardinality:
 * $X = \set {x_1, \dots, x_n, x_{n + 1} }$

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

By definition of cardinality:
 * $\size {\set {x_1, \dots, x_n} } = n$

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

Then by Induction Hypothesis:
 * $\map P {\set {x_1, \dots, x_n} }$

By definition of subset:
 * $x_{n + 1} \in A$

Thus by assumption:
 * $\map P X$

By the Principle of Mathematical Induction:
 * $\forall X \subseteq A: \paren {\size X = \size A \implies \map P X}$

Hence:
 * $\map P A$