1+1 = 2/Proof 2

Proof
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 $n = 0$:

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.