Can you multiply (and in general, commutatively-operate) equations together? Seems like a high-school question, but I don't have an extensive background in abstract algebra I actually didn't come across a technically rigorous proof  of "why" one is allowed to add completely separate equations together.
Well, in addition to adding equations together, can you multiply equations together such that for $x = y$ and $z = t,$ their product is $xy = zt$? Or is it a different result under some other more conventional equation-operation?
And not only would I ask if you can multiply them together, but let's suppose you have an arbitrary commutable operation defined over functions of a complex variable. Can you commutative-operation equations together in the general (or almost general) case? Something is missing from my knowledge about strcture-preserving operations acting on equivalence relations.
 A: It's not really proper to talk about multiplying equations; rather, we're applying the basic laws of equality, in particular reflexivity and substitution, to appropriate expressions.
Specifically, let's suppose we are assuming $$x=y\quad\mbox{and}\quad z=t.$$ Now by reflexivity we also know $$xz=xz.$$ Note that this has nothing to do with our assumptions above or the particular properties of multiplication; any expression is always equal to itself.
Now we combine the three equations we have - our two nontrivial assumptions and our trivial consequence of reflexivity - using the principle of substitution:

(Substitution) If $\varphi$ is a true statement and $a=b$ is a true equation, then $\varphi'$ is also a true statement whenever $\varphi'$ is gotten from $\varphi$ by replacing some instances of "$a$" with "$b$."

Here we apply this as follows:

*

*From $xz=xz$ and $x=y$ we get $xz=yz$.


*From $xz=yz$ and $z=t$ we get $xz=yt$.
Note that in each application we've only substituted one of the possible two occurrences of the "old" term ($x$ and $z$ respectively). Also note that no particular properties of multiplication have been used here; the same strategy would work for exponentiation instead (showing that from $x=y$ and $z=t$ we can deduce $x^z=y^t$), even though exponentiation is not commutative (or associative for that matter).
