a + (b + 1) = (a + b) + 1,
which, apart from the difference of notation, is nothing but the equality (1), by which I have just defined addition.
Supposing the theorem true 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.