#1**+5 **

The proof starts from the Peano Postulates, which define the natural numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.

P2. If x is in N, then its "successor" x' is in N.

P3. There is no x such that x' = 1.

P4. If x isn't 1, then there is a y in N such that y' = x.

P5. If S is a subset of N, 1 is in S, and the implication (x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:

Def: Let a and b be in N. If b = 1, then define a + b = a' (using P1 and P2). If b isn't 1, then let c' = b, with c in N (using P4), and define a + b = (a + c)'.

Then you have to define 2:

Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1. Then 1 + 1 = 1' = 2

**Q.E.D.**

Guest Mar 7, 2017