# Induction of Finite Set

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## Contents

## Theorem Scheme

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:

- $\map P X$

$\Box$

By the Principle of Mathematical Induction:

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

Hence:

- $\map P A$

$\blacksquare$

## Sources

- Mizar article FINSET_1:sch 2