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.
Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.
You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:
Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.
