Logical fallacy? Suppose I have two mathematical statements : $A$ and $B$.
Suppose that $A$ is an already proven theorem.
Suppose that, to prove $B$, I use $A$ somewhere in the proof.
Therefore, $B$ is a proven theorem itself.
Now, suppose I want to find another proof for $A$. Specifically, I prove $A$ by resorting to $B$ somewhere in the proof.
Can I say that I have produced another valid proof for $A$ ?
 A: The name for this fallacy is begging the question/petitio principii.
A: Short answer: Not really.
Long answer: Not really. If you want to be pedant, it's a "new proof": you prove that $A$ is true, then you prove something else, and then you conclude that $A$ is true... instead of concluding that $A$ is true directly. It's as if you took a proof, inserted somewhere "therefore $1=1$", and then went on. It's a "new" proof, from a formal point of view, but any mathematician would (rightly) say it's not.
In other words, you have some set of axioms $T$. You can find deductions in a formal system that $T \vdash A$, $T \cup \{A\} \vdash B$ and $T \cup \{B\} \vdash A$. Then you can put them back to back (using modus ponens) to deduce something like $T \vdash ((A \wedge (A \implies B) \wedge (B \implies A)) \implies A)$. But in doing so, you're simply proving $A \implies A$.
A: Let's say our only axiom is A: (p$\rightarrow$(q$\rightarrow$(r$\rightarrow$q))) and we have uniform substitution and detachment as our only rules of inference.  We can obtain B: (r$\rightarrow$(s$\rightarrow$r)) as follows:
axiom 1           1 (p→(q→(r→q)))
by axiom 1        2 ((p→(q→(r→q)))→(r→(s→r)))
detachment 2 ,1   3 (r→(s→r))

To prove B (r$\rightarrow$(s$\rightarrow$r)) we used A (p→(q→(r→q))) in the proof.  Now let's prove A by restoring to B somewhere in the proof as follows.
theorem             1 (r→(s→r))
1 r/(r→(s→r)), s/q  2 ((r→(s→r))→(q→(r→(s→r))))
detachment 2, 1     3 (q→(r→(s→r)))
3 q/p, r/q, s/r     4 (p→(q→(r→q)))

In this case you have produced another valid proof for A (the original proof of A consisted of listing it as an axiom).  So, the answer to your question is "yes" in certain systems of logic.
