Max Operation Preserves Total Ordering

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Theorem

Let $\struct {S, \preceq}$ be a totally ordered set.

Let $a, b, c, d \in S$:

$a \preceq b, c \preceq d$

Then:

$\max \set{a, c} \preceq \max \set{b, d}$

where $\max$ denotes the max operation on $\struct {S, \preceq}$.

Proof

From Max Operation Equals an Operand, either:

$\max \set{a, c} = a$

or

$\max \set{a, c} = c$

Without loss of generality, suppose:

$\max \set{a, c} = a$

We have:

\(\ds \max \set{a, c}\)

\(=\)

\(\ds a\)

by hypothesis

\(\ds \)

\(\preceq\)

\(\ds b\)

by hypothesis

\(\ds \)

\(\preceq\)

\(\ds \max \set{b, d}\)

Max Operation Yields Supremum of Parameters and Definition of Supremum

By transitivity of an ordering:

$\max \set{a, c} \preceq \max \set{b, d}$

$\blacksquare$