1+1 = 2/Proof 2

Proof
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 $n = 0$:

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:

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


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

Hence the result.