Proof of Zero Products and Elementary Identities in a Field

All axiom numbers below refer to Field.

Uniqueness of additive inverses. If $u,v,w\in F$ satisfy $u+v=0$ and $u+w=0$, then, using axioms 2, 1 and 4,

In particular $-u$ is the only element $v$ with $u+v=0$, and since $(-u)+u=0$ by axioms 3 and 4, we get $-(-u)=u$.

Right distributivity. For all $p,q,s\in F$, axioms 8 and 9 give

Claim 1. By axiom 2, $0+0=0$, so axiom 9 gives

Adding $-(x0)$ to both sides and using axioms 1, 3 and 2 gives

By axiom 8, $0x=x0=0$.

Claim 2. By right distributivity, axiom 3 and claim 1,

so $(-x)y$ is an additive inverse of $xy$, and by the uniqueness recorded above $(-x)y=-(xy)$. Applying this twice, together with axiom 8,

Claim 3. Suppose $xy=0$ and $x\ne0$. By axiom 7 there is $x^{-1}\in F$ with $xx^{-1}=1$, and $x^{-1}x=1$ by axiom 8. Using axioms 6, 8, 5 and claim 1,

Hence $x=0$ or $y=0$.

Claim 4. By right distributivity, axiom 9, claim 2 and axiom 8,

Since $yx=xy$ by axiom 8, the middle terms satisfy $xy+\bigl(-(xy)\bigr)=0$, so axioms 1, 4, 3 and 2 give

Claim 5. By right distributivity, axiom 9 and axiom 8,

the last step using $yx=xy$ together with axioms 1 and 4.