Prove that the order on integers is antisymmetric

Question

Let $m, n \in \mathbb{Z}$. If $m \le n \le m$ then $m = n$.

The definition of order is: if $m>n$ it means $m-n \in\mathbb{N}$.

I know that by definition this means either $m<n$ and $n<m$ or $m=n$, but I'm having trouble proving $m=n$.

Answer 1

You are given $(m \lt n) \vee (m=n)$, so assume $m \lt n$. Can you show $n \not \lt m$?

Answer 2

By assuming $n\neq m$, you have either $n<m$ or $m<n$. Then assume one of it, say $n<m$, together with $n\geq m$. Can you get a contradiction?

Answer 3

$m\leq n\leq m$ implies:

$1.$ $m$ is less than or equal to $n$

$2.$ $m$ is greater than or equal to $n$

Taking the intersection of the above two conditions, we get $m=n$.

Answer 4

Let's just start at the most complex side and do what comes naturally: \begin{align} & m \le n \;\land\; n \le m \\ \equiv & \qquad \text{"basic property of $\;\le\;$, twice -- to introduce $\;=\;$ as in our goal"} \\ & (m < n \lor m = n) \;\land\; (n < m \lor n = m) \\ \equiv & \qquad \text{"logic: factor out $\;m = n\;$ using distribution"} \\ & (m < n \;\land\; n < m) \lor m = n \\ \equiv & \qquad \text{"... -- to get rid of the left hand part"} \\ & \ldots & \end{align} Can you finish this calculation, given the rules you know about $\;<\;$?