# Using a conjecture to solve a conjecture

Is it mathematically correct to use a conjecture to prove another conjecture ? And if the second one is proved, does that mean that we'll only focus on proving the first ? There are many cases that emerge from this.

This idea came to my mind while studying Goldbach's Conjecture

• The proof would be a simple proof that "Conjecture $A \implies$ Conjecture $B$". This is totally valid. The only invalid thing to do is to claim that this constitutes an unconditional proof of Conjecture $B$. But there's quite a bit of work out there that does this - many papers have been published proving results in number theory are true assuming the Riemann Hypothesis, for instance. Commented Oct 4, 2021 at 3:30
• There are probably hundrets or even thousands of papers that start with "Suppose Riemann's Hypothesis is true", but then the results do not say anything about the truth of the Riemann Hypothesis, and are first valid once you have proven this conjecture. So while you can assume anything at first, it might turn out to be pointless or not, once the conjecture is proven or disproven. Commented Oct 4, 2021 at 3:31
• A famous example is when Ribet (after Hellegouarch, Frey, Serre) showed in 1986 that the Taniyama-Shimura conjecture implies Fermat's Last Theorem, see en.m.wikipedia.org/wiki/Ribet%27s_theorem. This inspired Wiles to prove the Taniyama-Shimura conjecture, thereby proving Fermat a decade later. Commented Oct 4, 2021 at 14:52

A mathematical theorem is typically a statement $$P\implies Q$$ that is logically true under the theory's axioms.

A conjecture is an unproven theorem.

Consider the following valid mathematical proof:
$$P\quad$$ and $$\quad P\implies (A\implies B);$$
therefore $$\quad A\implies B.$$
The first line contain the assumptions (also called premises or hypotheses), while the second line contains the conclusion.

• If we know $$P$$ to be true, and $$\big[P\implies (A\implies B)\big]$$ to be a theorem, then we can rightly consider $$\big[A\implies B\big]$$ to be another theorem. In this case, the above proof is a sound argument.

• On the other hand, if we know $$P$$ to be true, but $$\big[P\implies (A\implies B)\big]$$ is merely a conjecture, then $$\big[A\implies B\big]$$ has to be another conjecture.

1. If the first conjecture becomes proven, then we're back to the previous case, and the second conjecture automatically too becomes proven.
2. But if the first conjecture becomes disproven, then the above proof, while still valid, becomes unsound, and the second conjecture remains a conjecture. (Even though the second conjecture had been derived from the first, disproving the latter does not automatically disprove the former.)