Simple but challenging probability identity I am reading E.T. Jayne's Probability Theory book, and I am stuck on a "trivial" fact that he mentions at the end of the derivation of the sum rule in Chapter 2 (page 33):
He states that it is a trivial fact that given
$$
\begin{align}
w(AB|C) = w(A|BC)w(B|C) = w(B|AC)w(A|C) \\
w(A\bar{B}|C) = w(A|C)w(\bar{B}|AC)
\end{align}
$$
one can show that
$$
w^{m}(A|C) - w^m(A\bar{B}|C) = w^m(B|C) - w^m(B\bar{A}|C)
$$
where w is a probability measure.
Can someone demonstrate how this result follows?
Note, I think it's clear in the case that $m = 1$, both sides of the final equation are equal to $w(AB|C)$. I just don't see how you can derive this from the given equations for all powers of $m$.
 A: Suppose $x,y,z,w$ are complex numbers and $x^m - y^m = z^m-w^m$ for both $m=1$ and $m=2$.
Then I claim either $x=y$ and $z=w$, or $x=z$ and $y=w$.
Proof: $$x^2 - y^2 - z^2 + w^2 = 2(w-y)(w-z) + (x-y-z+w)(x+y+z-w)$$
so $(w-y)(w-z) = 0$, i.e. $w=y$ or $w=z$.  If $w=y$ we must have $x=z$ and if $w=z$ then $x=y$.
Thus in your case, it is impossible to have your equation true for both $m=1$ and $m=2$, let alone all positive integers $m$, unless either $w(A\mid C) = w(B \mid C)$ and $w(A \overline{B}\mid C) = w(B\overline{A} \mid C)$, or $w(A \mid C) = w(
A \overline{B} \mid C)$ and $w(B \mid C) = w(B \overline{A} \mid C)$.
A: This works for $m=1$. I don't know if this can be extended to $m \neq 1.$
Notice that
$w(A|C) - w(A\bar{B}|C) = w(B|C) - w(B\bar{A}|C) \Rightarrow \\
w(AC)w(C) - w(A\bar{B}C)w(C) = w(BC)w(C) - w(B\bar{A}C)w(C) \Rightarrow \\
w(AC) - w(A\bar{B}C) = w(BC) - w(B\bar{A}C) \Rightarrow \\
w(AC) + w(B\bar{A}C) = w(A\bar{B}C) + w(BC).$
Observe that $AC$ and $B\bar{A}C$ are disjoint sets. Therefore:
$w(AC) + w(B\bar{A}C) = w(AC+ B\bar{A}C)$.
Similarly, $BC$ and $A\bar{B}C$ are disjoint, and:
$w(BC) + w(A\bar{B}C) = w(BC+ A\bar{B}C)$.
Moreover, using DeMorgan theorem:
$$AC + B\bar{A}C = C(A + B\bar{A}) = C\overline{(\bar{A}\overline{B\bar{A}})} = \\
= C\overline{(\bar{A}(\bar{B} + A))} = C\overline{(\bar{A}\bar{B})} = C(A+B).$$
Similarly:
$$BC+ A\bar{B}C = \text{proof left to the reader} = C(A+B).$$
