Proof of Additive Cancellation and Elementary Additive Identities in a Field

Throughout we use only the addition axioms of a field: addition is associative and commutative, the additive identity satisfies $x+0=x$ for every $x$, and every $x$ satisfies $x+(-x)=0$. By commutativity these also give $0+x=x$ and $(-x)+x=0$.

Claim 1. Suppose $x+y=0$. Then

Since $x+y=0$ gives $y+x=0$ by commutativity, the same computation with the roles of $x$ and $y$ interchanged gives $x=-y$.

Claim 2. Suppose $x+z=y+z$. Then

Claim 3. The additive inverse axiom gives $x-x=x+(-x)=0$. Suppose now that $x-y=0$, that is, $x+(-y)=0$. Then

Conversely, if $x=y$, then $x-y=x-x=0$ by what has just been proved.

Claim 4. Since $0+0=0$, claim 1 applied to the pair $0,0$ gives $0=-0$. Hence $x-0=x+(-0)=x+0=x$, and $0-x=0+(-x)=-x$.

Claim 5. Since $(-x)+x=0$, claim 1 applied to the pair $-x,x$ gives $x=-(-x)$.

Claim 6. By associativity and commutativity of addition,

so claim 1 gives $(-x)+(-y)=-(x+y)$. Applying this identity to $x$ and $-y$ in place of $x$ and $y$, and then using claim 5 and commutativity,