From the two field axioms and additive identity properties provided, I've given a proof of (2)$(\forall x)(x \in \mathbb{F} \rightarrow x\cdot e = e \cdot x = e)$. I wanted to know if the proof is correct and/or the proof needs improvements?
Field Axioms (A4) and (M1):
(A4) $(\exists e)[e \in \mathbb{F} \land (\forall x)(x \in \mathbb{F} \rightarrow x + e = e + x = x)]$ (additive identity property)
(M1) $ \bullet: \mathbb{F} \times \mathbb{F} \to \mathbb{F}$ (closure under multiplication)
Properties for the Additive Identity:
(1) $(\forall x)(x \in \mathbb{F} \land x+x = x \rightarrow x=e)$
(2) $(\forall x)(x \in \mathbb{F} \rightarrow x\cdot e = e \cdot x = e)$
proof of (2).
Since $e \in \mathbb{F}$, by (A4), we have that $e+e=e$. Now for any $x \in \mathbb{F}$, $$x\cdot(e+e) = x\cdot e$$ By the Distributive Law, $$x\cdot e + x\cdot e = x\cdot e$$ Since $x\cdot e \in \mathbb{F}$ by (M1), (1) implies that $x\cdot e = e$. In a similar process, we show that $e \cdot x = e$ using the fact that $(e + e)\cdot x = e \cdot x$. QED.