@. Case, Theorem 5 implies that low(n ? 1) > 3, and in particular n ? 1 isn't 3-odd. So the first IH is G(n ? 1) = G(n ? 1) By theorem 15, we also know that G(n ? 1) = G(n) ? 1. So G(n ? 1) + 1 = G(n ? 1) + 1 = G(n) By theorem 17, we know that G(n) has an even lowest rank, so it cannot be 3-odd, and the second IH is G(G(n ? 1) + 1) = G(G(n ? 1) + 1) Since low(G(n)) is even, we can use theorem 15 once more, n) ? 1) = G(G(n)) ? 1. Finally: G(G(n ? 1) + 1) = G(G(n ? 1) + 1) = G(G(n)) = 1 + G(G(n ? 1)

@. Case, By theorem 5, low(n ? 1) = 2 hence the first IH is G(n ? 1) = G(n ? 1) We also know that G(n ? 1) = G(n) by theorem 15, and that low(G(n ? 1) + 1) = low(G(n) + 1) is even by theorem 17. The second IH is hence, n?1)+1) = G(G(n?1)+1

@. Case, is 3-odd (theorem 6), so the first IH is G(n ? 1) = 1 + G(n ? 1) We also know that G(n ? 1) = G(n) ? 1 by theorem 15 By theorem 16 we know that low(G(n)) = low(n) ? 1 > 3, so by theorem 4 low(G(n) + 1) = 2, and G(n ? 1) + 1 = 1 + G(n) cannot be 3-odd : the second IH is hence G(G(n ? 1) + 1) = G(G(n ? 1) + 1) Moreover low(G(n)) is also known to be odd, theorem 15 we have G(G(n ? 1)) = G(G(n) ? 1) = G(G(n)). The final step is G(1+ G(n)) = 1+ G(G(n)) also by theorem 15