Lemma

— Let $P \in \mathscr{M}$ and $x \notin P$; then there exists $P' \in \mathscr{M}$ such that $P \subset P'$ and $-x \in P'$.

Take $P' = P - xP$, and check that $P'$ has the required properties. Since $0 \in C \subset P$, we have $P \subset P'$. Whence $C \subset P'$ and $K_+ \subset P'$. Since $1 \in C \subset P$ we have $-x \in P'$. We have

whence (AP_I). We have whence (AP_{II}). Finally, let us check (AP_{III}): suppose given an identity of the form $p - xq = -(r - xs)$ where $p, q, r, s$ belong to P; we deduce from this the relation $x(s + q) = p + r$; if $s + q \neq 0$ we have contrary to the hypothesis; hence $s + q = 0$, whence $p + r = 0$; since P satisfies (AP_{III}) we deduce that $s = q = r = p = 0$, which completes the proof.