# 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 $a = b$.

Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:

- $a + c = b + c$

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

Similarly, suppose $b < a$.

Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:

- $b + c < a + c$

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 $b = c$.

Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:

- $a + b = a + c$

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

Similarly, suppose $c < b$.

Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:

- $a + c < a + b$

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 $<$.

From Natural Number Addition is Commutative and Right Cancellable Commutative Operation is Left Cancellable:

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

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

Hence the result.

$\blacksquare$

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): $2.2$: Theorem $2.14 \ \text{(i)}$