for c = γ, I say it will be true for c = γ + 1.
In fact, supposing
(a + b) + γ = a + (b + γ),
it follows that
[(a + b) + γ] + 1 = [a + (b + γ)] + 1
or by definition (1)
(a + b) + (γ + 1) = a + (b + γ + 1) = a + [b + (γ + 1)],
which shows, by a series of purely analytic deductions, that the theorem is true for γ + 1.
Being true for c = 1, we thus see successively that so it is for c = 2, for c = 3, etc.
Commutativity.—1º I say that
a + 1 = 1 + a.
The theorem is evidently true for a = 1; we can verify by purely analytic reasoning that if it is true for a = γ it will be true for a = γ + 1; for then
(γ + 1) + 1 = (1 + γ) + 1 = 1 + (γ + 1);
now it is true for a = 1, therefore it will be true for a = 2, for a = 3, etc., which is expressed by saying that the enunciated proposition is demonstrated by recurrence.
2º I say that
a + b = b + a.