Cox proof of product rule - step explanation

Question

I'm going through "Probability Theory - The logic of science" written by E.T. Jaynes and I have a problem with one step on page 27/28 in the proof of the product rule. The idea here is that we have a function $F$ for which: $$(AB|C)=F[(B|C), (A|BC)]$$ We want to proof its associativity: $$F[F(x, y), z] = F[x, F(y,z)]$$ We assume that F is differentiable. We denote: $$u = F(x,y)$$ $$v = F(y,z)$$ $$F_1(x,y) = \frac{\partial F}{\partial x}$$ $$F_2(x,y) = \frac{\partial F}{\partial y}$$ With this assumptions, we want to prove: $$F(x, v) = F(u, z)$$ We differentiate w.r.t. x and y: $$F_1(x,v)=F_1(u,z)F_1(x,y)$$ $$F_2(x,v)F_1(y,z)=F_1(u,z)F_2(x,y)$$ We then use: $$G(x,y)=\frac{F_2(x,y)}{F_1(x,y)}$$ And get equation E1: $$G(x,v)F_1(y,z)=G(x,y)$$ Which can be rewritten as equation E2: $$G(x,v)F_2(y,z)=G(x,y)G(y,z)$$ Now there's the part is hard for me to understand. We differentiate E1 w.r.t. z and E2 w.r.t. y. The proof states that left hand sides of both derivatives are the same. How is that? When I'm using product rule of calculus and chain rule, I get summation in E1 left side: $$G_2(x,v)F_2(y,z)F_1(y,z)+G(x,v)F_{12}(y,z)$$ and for equation E2, left side equals: $$G_2(x,v)F_1(y,z)F_2(y,z)+G(x,v)F_{21}(y,z)$$ So there's the quation - where's my mistake? I can see that first product of both equations derivative are the same but how this equations are the same? I can see two options: first, I shouldn't use product rule here (why?), second, for some reason $$F_{12}(y,z)=F_{21}(y,z)$$ but I can't see reason for any of these. I've checked original Cox proof in his book "The algebra of probable inference" but it's quite the same as in Jaynes book.

Answer 1

Yes, the double derivatives are the same whatever the performance order.

$$\begin{align}F_{12}(s,t) ~=~& \frac{\mathrm d ~}{\mathrm d s}\frac{\mathrm d ~}{\mathrm d t}F(s,t) \\[1ex] =~& \frac{\mathrm d ~}{\mathrm d t}\frac{\mathrm d ~}{\mathrm d s}F(s,t) \\[1ex]=~& F_{21}(s,t)\end{align}$$