Addition on 1-Based Natural Numbers is Cancellable

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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.

$\blacksquare$


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