Equivalence of Definitions of Minimal Element

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $T \subseteq S$ be a subset of $S$.

Definition 1 implies Definition 2
Let $x$ be an minimal element by definition 1.

That is:
 * $(1): \quad \forall y \in T: y \preceq x \implies x = y$

Aiming for a contradiction, suppose that:
 * $\exists y \in T: y \prec x$

Then by definition:
 * $y \preceq x \land x \ne y$

which contradicts $(1)$.

Thus by Proof by Contradiction:
 * $\nexists y \in T: y \prec x$

That is $x$ is a minimal element by definition 2.

Definition 2 implies Definition 1
Let $x$ be a minimal element by definition 2.

That is:
 * $(2): \quad \nexists y \in T: y \prec x$

Aiming for a contradiction, suppose that:
 * $\exists y \in T: y \preceq x: x \ne y$

That is:
 * $\exists y \in T: y \prec x$

which contradicts $(2)$.

Thus:
 * $\forall y \in T: y \preceq x \implies x = y$

Thus $x$ is a minimal element by definition 1.