1+1 = 2

Theorem
Define $0$ as the only element in the set $P\setminus s(P)$, where $s(P)$ is the image of the mapping $s$ defined in Peano structure, and $\setminus$ denotes the set difference.

Then define $1$ as the successor of $0$, and $2$ as the successor of $1$.

The theorem to be proven is:
 * $1 + 1 = 2$

Proof 1
Define $1$ as $s(0)$ and $2$ as $s(s(0))$.

Therefore the statement to be proven becomes:


 * $s(0)+s(0)=s(s(0))$

Thus:

Proof 2
Defining $1$ as $s(0)$ and $2$ as $s(s(0))$ the statement to prove becomes:


 * $s(0) + s(0) = s(s(0))$

By the definition of addition:


 * $\forall m \forall n: m + s(n) = s(m + n)$

Letting $m = s(0)$ and $n = 0$:

By the definition of addition:


 * $\forall m: m + 0 = m$

Letting $m = s(0)$:


 * $s(0) + 0 = s(0)$

Taking the successor of both sides:

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


 * $s(0) + s(0) = s(s(0))$

Hence the result.