# 1+1 = 2

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## 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 $s \left({0}\right)$ and $2$ as $s \left({s \left({0}\right)}\right)$.

Therefore the statement to be proven becomes:

- $s \left({0}\right) + s \left({0}\right) = s \left({s \left({0}\right)}\right)$

Thus:

\(\displaystyle \forall n \in P: \ \ \) | \(\displaystyle m + s \left({n}\right)\) | \(=\) | \(\displaystyle s \left({m + n}\right)\) | Definition of Addition | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle m + s \left({0}\right)\) | \(=\) | \(\displaystyle s \left({m+0}\right)\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \forall m \in P: \ \ \) | \(\displaystyle m + s \left({0}\right)\) | \(=\) | \(\displaystyle s \left({m + 0}\right)\) | |||||||||

\(\text {(1)}: \quad\) | \(\displaystyle \implies \ \ \) | \(\displaystyle s \left({0}\right) + s \left({0}\right)\) | \(=\) | \(\displaystyle s \left({s \left({0}\right) + 0}\right)\) | |||||||||

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

\(\text {(2)}: \quad\) | \(\displaystyle \implies \ \ \) | \(\displaystyle s \left({0}\right) + 0\) | \(=\) | \(\displaystyle s \left({0}\right)\) | |||||||||

\(\displaystyle s \left({s \left({0}\right) + 0}\right)\) | \(=\) | \(\displaystyle s \left({s \left({0}\right)}\right)\) | taking the successor on both sides of $(2)$ | ||||||||||

\(\text {(3)}: \quad\) | \(\displaystyle \implies \ \ \) | \(\displaystyle s \left({0}\right) + s \left({0}\right)\) | \(=\) | \(\displaystyle s \left({s \left({0}\right) + 0}\right)\) | from $(1)$ | ||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle s \left({s \left({0}\right)}\right)\) | 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 $n = 0$:

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

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:

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

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

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

Hence the result.

$\blacksquare$