Proof of Elementary Properties of the Minimum of Two Elements

Write $m=\min\{a,b\}$ and $M=\max\{a,b\}$. Throughout we use the reflexivity, antisymmetry, transitivity, and comparability axioms of a total order. By comparability, either $a\le b$ holds or it fails; we treat the two cases separately, and in each case the value of $m$ is read off from Minimum of Two Elements of a Totally Ordered Set and the value of $M$ from Maximum of Two Elements of a Totally Ordered Set.

Case 1: $a\le b$. Then $m=a$ and $M=b$.

1. Reflexivity gives $a\le a$, so $m\le a$; and $m=a\le b$ by the case hypothesis.

2. $m=a$.

3. Suppose $c\le m$. Then $c\le a$, and transitivity with $a\le b$ gives $c\le b$. Conversely, if $c\le a$ and $c\le b$, then in particular $c\le a=m$.

4. Apply Minimum of Two Elements of a Totally Ordered Set to the ordered pair $(b,a)$. If $b\le a$, then $\min\{b,a\}=b$; combining $b\le a$ with the case hypothesis $a\le b$, antisymmetry gives $a=b$, so $\min\{b,a\}=b=a=m$. If instead $b\le a$ fails, then $\min\{b,a\}=a=m$.

5. $m=a\le b=M$ by the case hypothesis.

Case 2: $a\le b$ fails. By comparability, $b\le a$. Then $m=b$ and $M=a$.

1. $m=b\le a$, and reflexivity gives $b\le b$, so $m\le b$.

2. $m=b$.

3. Suppose $c\le m$. Then $c\le b$, and transitivity with $b\le a$ gives $c\le a$. Conversely, if $c\le a$ and $c\le b$, then in particular $c\le b=m$.

4. Since $b\le a$, applying Minimum of Two Elements of a Totally Ordered Set to the ordered pair $(b,a)$ gives $\min\{b,a\}=b=m$.

5. $m=b\le a=M$.

In both cases all five claims hold, which completes the proof.