Q:

Need help understanding the logical structure behind solving an equation?

Suppose I have an equation $f(x) = 0$, and I know the solution exists. The solution is $x = a$ (I don't know the actual solution to begin with, I just know it exists). I perform a series of $irreversible$ operations and arrive at the answer:
$f(x) = 0 \implies \cdots = \cdots \implies \cdots = \cdots \implies x = a$
which is just a big $f(x) = 0 \implies x = a$.
Since the operations are irreversible, we cannot go in the other direction and claim $ x = a  \implies f(x) = 0$
Nevertheless, we know that the solution is $a$, so if we plug in $a$ into $f$ we get zero. But hold on a minute, the statement "if we plug in $a$ into $f$ we get zero" really means $x = a \implies f(x) = 0$; we just went in the other direction, which is something we said we couldn't do!
What is wrong here? Does the fact that $f(x) = a$ has a solution imply that there must be a reversible series of steps between $f(x) = a$ and $x = a$?
Edit: A condensed version
1) We know $f(x) = 0$ has a solution. 
2) We can prove $f(x) = 0 \implies x = a$ using irreversible steps. 
3) Now that we computed what $a$ actually is, by 1) we know that $x = a \implies f(x) = 0$ is true. Does it follow now that there exists another method to solve the equation using purely reversible steps?

A:

This is impossible to answer in any rigorous way without having a precise definition of the sort of "steps" you want to consider.  But basically, the answer is no.
It happens all the time in math that we can separately prove $P\Rightarrow Q$ and $Q\Rightarrow P$, but there is no direct way to prove $P\Leftrightarrow Q$ in both directions at once.  So there is no reason to expect that just because you can prove both $f(x)=0\Rightarrow x=a$ and $x=a\Rightarrow f(x)=0$, you can prove $f(x)=0\Leftrightarrow x=a$ by a series of reversible steps.  After all, any mathematical statement at all can be encoded in equations like this by defining your function $f$ appropriately.

