# 1+1 = 2

Jump to navigation
Jump to search

## Theorem

Define $0$ as the only element in the set $P \setminus s \left({P}\right)$, where:

- $P$ is the Peano Structure
- $s \left({P}\right)$ is the image of the mapping $s$ defined in Peano structure
- $\setminus$ denotes the set difference.

The theorem to be proven is:

- $1 + 1 = 2$

where:

- $1 := s \left({0}\right)$
- $2 := s \left({1}\right) = s \left({s \left({0}\right)}\right)$
- $+$ denotes addition
- $=$ denotes equality
- $s \left({n}\right)$ denotes the successor function as defined by Peano

## Proof 1

$1$ is defined as $\map s 0$ and $2$ as $\map s {\map s 0}$.

Therefore the statement to be proven becomes:

- $\map s 0 + \map s 0 = \map s {\map s 0}$

Thus:

\(\ds \forall n \in P: \, \) | \(\ds m + \map s n\) | \(=\) | \(\ds \map s {m + n}\) | Definition of Addition | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds m + \map s 0\) | \(=\) | \(\ds \map s {m + 0}\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \forall m \in P: \, \) | \(\ds m + \map s 0\) | \(=\) | \(\ds \map s {m + 0}\) | ||||||||||

\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map s 0 + \map s 0\) | \(=\) | \(\ds \map s {\map s 0 + 0}\) | ||||||||||

\(\ds \forall m \in P: \, \) | \(\ds m + 0\) | \(=\) | \(\ds m\) | Definition of Addition | ||||||||||

\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map s 0 + 0\) | \(=\) | \(\ds \map s 0\) | ||||||||||

\(\ds \map s {\map s 0 + 0}\) | \(=\) | \(\ds \map s {\map s 0}\) | taking the successor on both sides of $(2)$ | |||||||||||

\(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map s 0 + \map s 0\) | \(=\) | \(\ds \map s {\map s 0 + 0}\) | from $(1)$ | |||||||||

\(\ds \) | \(=\) | \(\ds \map s {\map s 0}\) | from $(2)$ |

$\blacksquare$

## Proof 2

Defining $1$ as $\map s 0$ and $2$ as $\map s {\map s 0}$, the statement to be proven becomes:

- $\map s 0 + \map s 0 = \map s {\map s 0}$

By the definition of addition:

- $\forall m \in P: \forall n \in P: m + \map s n = \map s {m + n}$

Letting $m = \map s 0$ and $n = 0$:

\(\text {(1)}: \quad\) | \(\ds \map s 0 + \map s 0\) | \(=\) | \(\ds \map s {\map s 0 + 0}\) |

By the definition of addition:

- $\forall m: m + 0 = m$

Letting $m = \map s 0$:

- $\map s 0 + 0 = \map s 0$

Taking the successor of both sides:

\(\text {(2)}: \quad\) | \(\ds \map s {\map s 0 + 0}\) | \(=\) | \(\ds \map s {\map s 0}\) |

Applying Equality is Transitive to $(1)$ and $(2)$ we have:

- $\map s 0 + \map s 0 = \map s {\map s 0}$

Hence the result.

$\blacksquare$