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
\(\displaystyle \implies \ \ \) \(\displaystyle m + s \left({0}\right)\) \(=\) \(\displaystyle s \left({m+0}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle \forall m \in P: \ \ \) \(\displaystyle m + s \left({0}\right)\) \(=\) \(\displaystyle s \left({m + 0}\right)\)
\((1):\quad\) \(\displaystyle \implies \ \ \) \(\displaystyle s \left({0}\right) + s \left({0}\right)\) \(=\) \(\displaystyle s \left({s \left({0}\right) + 0}\right)\)
\(\displaystyle \forall m \in P: \ \ \) \(\displaystyle m + 0\) \(=\) \(\displaystyle m\) Definition of Addition
\((2):\quad\) \(\displaystyle \implies \ \ \) \(\displaystyle s \left({0}\right) + 0\) \(=\) \(\displaystyle s \left({0}\right)\)
\(\displaystyle s \left({s \left({0}\right) + 0}\right)\) \(=\) \(\displaystyle s \left({s \left({0}\right)}\right)\) taking the successor on both sides of $(2)$
\((3):\quad\) \(\displaystyle \implies \ \ \) \(\displaystyle s \left({0}\right) + s \left({0}\right)\) \(=\) \(\displaystyle s \left({s \left({0}\right) + 0}\right)\) from $(1)$
\(\displaystyle \) \(=\) \(\displaystyle s \left({s \left({0}\right)}\right)\) from $(2)$

$\blacksquare$


Proof 2

Defining $1$ as $s \left({0}\right)$ and $2$ as $s \left({s \left({0}\right)}\right)$, the statement to be proven becomes:

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


By the definition of addition:

$\forall m \in P: \forall n \in P: m + s \left({n}\right) = s \left({m + n}\right)$

Letting $m = s \left({0}\right)$ and $n = 0$:

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


By the definition of addition:

$\forall m: m + 0 = m$

Letting $m = s \left({0}\right)$:

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

Taking the successor of both sides:

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


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

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

Hence the result.

$\blacksquare$