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I need help proving this theorem using the axioms and theorems shown in the picture: If $n,m$ are in the natural numbers and $n\ge m$ then $n-m$ is in the natural numbers (https://i.stack.imgur.com/I665o.jpg)

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    $\begingroup$ Please don't use pictures. $\endgroup$ – Dietrich Burde Nov 8 at 16:06
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    $\begingroup$ especially sideways ones $\endgroup$ – J. W. Tanner Nov 8 at 16:10
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  1. Since $n\geq m$, $n-m\geq 0$.
  2. Since $n,m$ are both integers (being natural numbers), $n-m$ is also an integer (because the integers are a closed set).

Combining statements $1$ and $2$ says that $n-m$ is a non-negative integer. Therefore it is a natural number.

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