Natural Number Addition is Cancellable for Ordering

Theorem

Let $\N$ be the natural numbers.

Let $<$ be the strict ordering on $\N$.

Let $+$ be addition on $\N$.

Then:

$\forall a, b, c \in \N_{>0}: a + c < b + c \implies a < b$

$\forall a, b, c \in \N_{>0}: a + b < a + c \implies b < c$

That is, $+$ is cancellable on $\N$ for $<$.

Proof

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