# 1+1 = 2

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## Theorem

Define $0$ as the unique element in the set $P \setminus \map s P$, where:

- $P$ is the Peano Structure
- $\map s P$ is the image of the mapping $s$ defined in Peano structure
- $\setminus$ denotes the set difference.

Then:

- $1 + 1 = 2$

where:

- $1 := \map s 0$
- $2 := \map s 1 = \map s {\map s 0}$
- $+$ denotes addition
- $=$ denotes equality
- $\map s n$ denotes the successor mapping.

## Proof 1

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

Hence the statement to be proven becomes:

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

Thus:

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

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

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

Then: | |||||||||||||||

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

\(\ds \leadsto \ \ \) | \(\ds \map s 0 + 0\) | \(=\) | \(\ds \map s 0\) | as $\map s 0 \in P$ | |||||||||||

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

\(\ds \leadsto \ \ \) | \(\ds \map s 0 + \map s 0\) | \(=\) | \(\ds \map s {\map s 0}\) | substituting for $\map s {\map s 0 + 0}$ from $(1)$ |

$\blacksquare$

## 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.

$\blacksquare$