# 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:

$\size X = 0$
$X = \O$

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:

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

By definition of cardinality:

$\size {\set {x_1, \dots, x_n} } = n$
$\set {x_1, \dots, x_n} \subseteq X \subseteq A$

Then by Induction Hypothesis:

$\map P {\set {x_1, \dots, x_n} }$

By definition of subset:

$x_{n + 1} \in A$

Thus by assumption:

$\map P X$

$\Box$

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

Hence:

$\map P A$

$\blacksquare$