Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing

Theorem
Let $\left({S, \preceq_1}\right)$ be a totally ordered set.

Let $\left({T, \preceq_2}\right)$ be an ordered set.

Let $\phi: S \to T$ be a mapping.

Then $\phi$ is a dual order embedding $\phi$ is strictly decreasing.

That is:


 * $\forall x, y \in S: x \preceq_1 y \iff \phi \left({y}\right) \preceq_2 \phi \left({x}\right)$


 * $\forall x, y \in S: x \prec_1 y \implies \phi \left({y}\right) \prec_2 \phi \left({x}\right)$

Forward Implication
Let $\phi$ be a dual order embedding.

Then $\phi$ is an order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \succeq_2}\right)$, where $\succeq_2$ is the dual of $\preceq_2$.

Thus by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing:
 * $\phi: \left({S, \preceq_1}\right) \to \left({T, \succeq_2}\right)$ is strictly increasing.

Thus:
 * $\forall x, y \in S: x \prec_1 y \implies \phi \left({x}\right) \succ_2 \phi \left({y}\right)$

so:
 * $\forall x, y \in S: x \prec_1 y \implies \phi \left({y}\right) \prec_2 \phi \left({x}\right)$

Thus $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly decreasing.

Reverse Implication
Suppose that $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly decreasing.

Then by the same argument as above:
 * $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is strictly increasing.

Thus by Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing, $\phi$ is an order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \succeq_2}\right)$.

So $\phi$ is a dual order embedding of $\left({S, \preceq_1}\right)$ into $\left({T, \preceq_2}\right)$.