Natural Number Multiplication is Cancellable

Theorem
Let $\N_{> 0}$ be the $1$-based natural numbers.

Let $\times$ be multiplication on $\N_{>0}$.

Then:
 * $\forall a, b, c \in \N_{>0}: a \times c = b \times c \implies a = b$
 * $\forall a, b, c \in \N_{>0}: a \times b = a \times c \implies b = c$

That is, $\times$ is cancellable on $\N_{>0}$.

Proof
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds:
 * $a = b$
 * $a < b$
 * $b < a$

Suppose $a < b$.

Then by Ordering on $1$-Based Natural Numbers Compatible with Multiplication:
 * $a \times c < b \times c$

By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a \times c = b \times c$.

Similarly, suppose $b > a$.

Then by Ordering on $1$-Based Natural Numbers Compatible with Multiplication:
 * $b \times c < a \times c$

By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a \times c = b \times c$.

The only other possibility is that $a = b$.

So
 * $\forall a, b, c \in \N_{>0}: a \times c = b \times c \implies a = b$

and so $\times$ is right cancellable on $\N_{>0}$.

From Natural Number Multiplication is Commutative and Right Cancellable Commutative Operation is Left Cancellable:
 * $\forall, b, c \in \N_{>0}: a \times b = a \times c \implies b = c$

So $\times$ is both right cancellable and left cancellable on $\N_{>0}$.

Hence the result.