Equivalence of Definitions of Injection/Definition 1 iff Definition 1 a

Proof
Let $f: S \to T$ be an injection by definition 1.

Thus:
 * $\forall x_1, x_2 \in S: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$

Let $x_1 \ne x_2$.

Aiming for a contradiction, suppose that:
 * $f \left({x_1}\right) = f \left({x_2}\right)$

Then by definition 1:
 * $x_1 = x_2$

This contradicts the assumption that $x_1 \ne x_2$.

Thus by Proof by Contradiction:
 * $\forall x_1, x_2 \in S: x_1 \ne x_2 \implies f \left({x_1}\right) \ne f \left({x_2}\right)$

Thus $f$ is an injection by definition 1 a.

Let $f: S \to T$ be an injection by definition 1 a.

Thus:
 * $\forall x_1, x_2 \in S: x_1 \ne x_2 \implies f \left({x_1}\right) \ne f \left({x_2}\right)$

Let $f \left({x_1}\right) = f \left({x_2}\right)$.

Aiming for a contradiction, suppose that:
 * $x_1 \ne x_2$

Then by definition 1 a:
 * $f \left({x_1}\right) \ne f \left({x_2}\right)$

This contradicts the assumption that $f \left({x_1}\right) = f \left({x_2}\right)$.

Thus by Proof by Contradiction:
 * $\forall x_1, x_2 \in S: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$

Thus $f$ is an injection by definition 1.