By the Origin Lemma, there is an h-inversion hP interchanging O and P.
Suppose that hP(Q) = R, so hP(R) = Q as hP has order 2.
Again by the Lemma, there is an h-inversion hR interchanging O and R.
Let t = hPohRohP.
Then a direct calculation shows that t(P) = Q and t(Q) = P.
Also, t is an h-inversion by the the algebraic inversion theorem.
If u is a second h-inversion interchanging P and Q, then a calculation
We make frequent use
of the fact that, for an
h-inversion h, we have
h-1 = h.