Addition on 1-Based Natural Numbers is Cancellable

Theorem

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

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 $\N_{>0}$.

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 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 $\N_{>0}$.

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

• 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 4$: The natural numbers: $\text{A} 3$
• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.12$