Proof of Elementary Properties of the Maximum of Two Elements

Throughout we use the axioms of a total order: reflexivity, antisymmetry, transitivity and comparability. By the definition of the maximum, exactly one of the following two cases occurs.

Case A: $a\le b$. Then $\max\{a,b\}=b$.

Case B: $a\le b$ does not hold. Then $\max\{a,b\}=a$, and comparability gives $b\le a$.

Claim 2. In Case A the maximum is $b$ and in Case B it is $a$, so in either case it is $a$ or $b$.

Claim 1. In Case A we have $a\le b=\max\{a,b\}$, and $b\le b=\max\{a,b\}$ by reflexivity. In Case B we have $a\le a=\max\{a,b\}$ by reflexivity, and $b\le a=\max\{a,b\}$ as noted above.

Claim 3. Suppose $\max\{a,b\}\le c$. By claim 1 we have $a\le\max\{a,b\}$ and $b\le\max\{a,b\}$, so transitivity gives $a\le c$ and $b\le c$. Conversely, suppose $a\le c$ and $b\le c$. By claim 2 the element $\max\{a,b\}$ is $a$ or $b$, and in either case $\max\{a,b\}\le c$.

Claim 4. In Case A we have $\max\{a,b\}=b$. If $b\le a$ also holds, then antisymmetry gives $a=b$, and $\max\{b,a\}$ is $a$ or $b$ by claim 2, hence equals $b$; if $b\le a$ does not hold, then by the definition of the maximum $\max\{b,a\}=b$. In either case $\max\{b,a\}=b=\max\{a,b\}$.

In Case B we have $\max\{a,b\}=a$ and $b\le a$, so by the definition of the maximum $\max\{b,a\}=a=\max\{a,b\}$.