Problem 16 Proof

16. Proposition: For all $x, y \in \mathbb{N}$, if $y < S(x)$, then $y \leq x$.
Proof: We invoke the contrapositive of proposition 15a. Suppose neither $S(x) < y$ nor $S(x) = y$; then it is not the case that $x<y$. By trichotomy, as proven in 15b, this is equivalent to saying that $y< S(x)$ proves either $y<x$ or $y=x$. $\blacksquare$

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