Confusion on proof strategy: steps are reversible There are two proofs I'm trying to compare. I asked about one of them here: Proof organization: proving the set of functions $f: \mathbb{R} \to \mathbb{R}$ is a direct sum of even and odd functions, but am posting a new question because that was a specific proof statement, while this one is more on strategy.
I'm trying to understand whether, in general, solving a system of equations yields that every step is reversible. In that linked question, I learned that this is not the case when trying to write $f$ as a unique sum of an even function $f_e$ and an odd function $f_o$. I solved a system of equations in $f_e$ and $f_o$, namely
\begin{align*}
f(x) & = f_e (x) + f_o (x) \\ 
f(-x) & = f_e (x) - f_o (x),
\end{align*}
and found formulas for $f_o$ and $f_e$, and I learned that the steps are not reversible.
I'm trying to compare this to Ahlfor's proof where he solves the for the quotient of two complex numbers. He takes $\alpha + i\beta, \delta + i \gamma$, where $\delta + i \gamma \neq 0$, and solves for $x + iy$, the quotient of $\alpha + i \beta$ and $\delta + i \gamma$ by multiplying out the right-hand side of
$$
\alpha + i \beta = (\delta + i \gamma)(x + iy),
$$
and equating real and imaginary components. The reason this works, I believe, is because every step is reversible, but he doesn't show both directions. I believe the statement he proves is "if this quotient exists here is its formula," but he doesn't that the quotient actually exists.
Here is my sense of what's going on. In Ahlfor's proof, it is more or less "obvious" that every step is reversible, whereas in the case of writing a function as a unique sum of an even and odd function, it is not so obvious and there are additional things to check: that $f_e$ and $f_o$ are even and odd, respectively, for example.
I would appreciate some help on parsing this. I find myself struggling often with whether I need to prove a statement in both directions for a fully rigorous proof.
 A: 
I'm trying to understand whether, in general, solving a system of equations yields that every step is reversible.

Unless otherwise specified, solving a system of equations typically involves a forward chain of reasoning, so it is not generally true that every step is reversible; even when solving just a single equation

*

*\begin{align}&x+1=\sqrt{25}\\\implies {}&(x+1)^2=25\\\implies {}&x=-6\;\text{or}\;x=4.\end{align}
Since only the second value satisfies the given equation, we have that
$$x+1=\sqrt{25}\iff x=4.$$
we have an irreversible step: $(x+1)^2=25\kern.6em\not\kern-.6em\implies x+1=\sqrt{25}.$ On the other hand, a given system is, by definition, indeed always equivalent to its solution; for example, $$x{+}y{=}10,\,x{-}y{=}4\iff(x,y){=}(7,3).$$

In that linked question, I learned that this is not the case when trying to write $f$ as a unique sum of an even function $f_e$ and an odd function $f_o$. I solved a system of equations in $f_e$ and $f_o$, namely
\begin{align*}
f(x) & = f_e (x) + f_o (x) \tag1\\ 
f(-x) & = f_e (x) - f_o (x),\tag2
\end{align*}
and found formulas for $f_o$ and $f_e$

To be clear, you had derived equation $(2)$ from equation $(1).$ Most of your steps are not reversible, even in the subsection where you simultaneously ‘solved’ $(1)$ and $(2).$ Thus, you've merely argued this

*

*$$∀g\,∀h\,\Big(Eg∧Oh∧Sgh\implies Pg∧Mh\Big)$$ If a pair of even and odd functions $g$ and $h$ sums to $f,$
then $g(x)=\dfrac{f(x) + f(-x)}2$ and $h(x)=\dfrac{f(x) - f(-x)}2.$
and not its converse

*

*$$∀g\,∀h\,\Big(Pg∧Mh\implies Eg∧Oh∧Sgh\Big)$$ If functions $g$ and $h$ are such that $g(x)=\dfrac{f(x) + f(-x)}2$ and $h(x)=\dfrac{f(x) - f(-x)}2,$
then $g$ is even, $h$ is odd, and $f=g+h.$
Fortunately, the latter is trivial after having done the former; then, you can finally conclude that $$∀g\,∀h\,\Big(Eg∧Oh∧Sgh\iff Pg∧Mh\Big).$$

I'm trying to compare this to Ahlfor's proof where he solves the for the quotient of two complex numbers. He takes $\alpha + i\beta, \delta + i \gamma$, where $\delta + i \gamma \neq 0$, and solves for $x + iy$, the quotient of $\alpha + i \beta$ and $\delta + i \gamma$ by multiplying out the right-hand side of
$$\alpha + i \beta = (\delta + i \gamma)(x + iy),$$
and equating real and imaginary components. The reason this works, I believe, is because every step is reversible, but he doesn't show both directions. I believe the statement he proves is "if this quotient exists here is its formula," but he doesn't that the quotient actually exists.

To be clear, the author is solving—not for the quotient of two numbers but—for $x$ and $y,$ to ultimately obtain the quotient of two given numbers $z_1$ and $z_2$ where the latter is nonzero. Throughout their working, reversibility is not required (as is typically the case in proofs and arguments† and equation-solving where the provisional answers are checked against the original equation and restriction on it), while the quotient's existence is being implicitly understood rather than assumed.
 † After all an argument technically is a (forward-)conditional statement of the form $$\text{<conjunction of premises>} \implies \text{<conclusion>}.$$
