# Addition on 1-Based Natural Numbers is Cancellable for Ordering

## Theorem

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

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

Let $+$ be addition on $\N_{>0}$.

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 $\N_{>0}$ 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 $a + c < b + c$.

Suppose $a = b$.

$a + c = b + c$

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

Similarly, suppose $b < a$.

$b + c < a + c$

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

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

So

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

and so $+$ is right cancellable on $\N_{>0}$ for $<$.

Let $a + b < a + c$.

Suppose $b = c$.

$a + b = a + c$

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

Similarly, suppose $c < b$.

$a + c < a + b$

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

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

So

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

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

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

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

Hence the result.

$\blacksquare$