Natural Number Multiplication is Cancellable for Ordering

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

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

Let $<$ be the strict ordering 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 $\N_{>0}$ for $<$.

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

Let $a \times c < b \times c$.

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 $\N_{>0}$ for $<$.

Let $a \times b < a \times c$.

Suppose $b = c$.

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

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

Similarly, suppose $c < b$.

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

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

The only other possibility is that $b < c$.

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

and so $\times$ is left cancellable on $\N_{>0}$ for $<$.

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

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

Hence the result.