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Base case: b = 0:

$S(a) + 0 = S(a + 0)$

$S(a) = S(a)$

IH: $S(a) + (b + k) = S(a + (b + k))$

And so I tried getting the LHS the same as the RHS and this is as far as I got before hitting a roadblock

=> $S(a) + (b + S(k))$

=> $S(a) + S(b + k)$ - Using (ii) from addition

I had originally thought that perhaps for example, S(a) + S(b) = S(a + b), and if this worked, I'd get the RHS outcome. Perhaps I am overlooking at a possible substitution?

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    I understand that this is an excercise from your book, but it would be helpful to see what the actual question was. It's quite hard to help complete your proof if we don't know what you're trying to prove here... – 2017-02-22

1 Answers 1

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Base case: $S(a) + 0 = S(a)$

Inductive hypothesis:

Suppose: $S(a) + b = S(a+b)$

We must show that $S(a) + S(b) = S(a + S(b))$ based on the inductive hypothesis.

$S(a) + S(b) = S(S(a) + b)$ by the definition of addition.

$S(S(a) + b) = S(S(a + b))$ by the inductive hypothesis.

$S(a + S(b)) = S(S(a+b))$ by the definition of addition.

$S(a) + S(b) = S(a + S(b))$