Equality of coefficients of total differentials

[Guest]

I have trouble with one common inference about total differentials. In undergraduate physical chemistry textbooks this inference is treated as self evident. I have asked this questions several times on other sites and didn't got satisfactory answer. I got answers like. "It is self evident". I spend a lot of time and effort trying to solve this. Can you reduce this question to more general baby example? It would be nice if someone could give proof that could easily be converted to Fitch notation proof. Also geometric proof would be nice.
Why it is valid to inference:
[tex]\begin{cases} dz=\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy\\ dz=Adx + B dy \end{cases}[/tex]
follows by
[tex]\begin{cases} A=\frac{\partial f}{\partial x} \\ B=\frac{\partial f}{\partial y} \end{cases}[/tex]?

P. S. I spent a lot of time learning predicate logic myself and I know all the rules of inference. I have masters degree at chemistry.

[Guest]

I would say that follows from the definition of "=".

What definition of "=" are you using?

[Guest]

From $dz= \left(\frac{\partial df}{\partial x}\right)dx+ \left(\frac{\partial df}{\partial y}\right)dy$'
and $dz= A dx+ B dy$ it follows immediately that
$\left(\frac{\partial df}{\partial x}\right)dx+ \left(\frac{\partial df}{\partial y}\right)dy= A dx+ B dy$.

I hope that much is clear! That is what "Guest" was saying.

To complete this, you need to require that x and y are independent variables.
Then it is possible to vary x while holding y constant so that dy= 0 and we have
$\left(\frac{\partial df}{\partial x}\right)dx= A dx$ so that $\frac{\partial f}{\partial x}= A$.

And we can vary y while holding x constant so that dx= 0 and we have
$\left(\frac{\partial df}{\partial y}\right)dy= B dy$.