Strictly Decreasing Mapping is Decreasing

Theorem
A mapping that is strictly decreasing is a decreasing mapping.

Proof
Let $$\left({S; \preceq_1}\right)$$ and $$\left({T; \preceq_2}\right)$$ be posets.

Let $$\phi: \left({S; \preceq_1}\right) \to \left({T; \preceq_2}\right)$$ be strictly decreasing.

Note that, from Strictly Precedes, $$x \preceq_1 y \iff x = y \or x \prec_1 y$$. So:

$$ $$ $$

This leaves us with:

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