1+1 = 2/Proof 2

Theorem
Define $0$ as the only element in the set $P \setminus s \left({P}\right)$, where:
 * $s \left({P}\right)$ 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
Defining $1$ as $s \left({0}\right)$ and $2$ as $s \left({s \left({0}\right)}\right)$ the statement to prove becomes:


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

By the definition of addition:


 * $\forall m \in \N_{>0} \forall n \in \N_{>0}: m + s \left({n}\right) = s \left({m + n}\right)$

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

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:

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.