Suppose there are two statements, $A$ and $B$ that are independent. As far as I know one needn't to prove $A$ or $B$ either, it is enough to generate $C = A \land B$, and then proving $C$ shows $A$ and $B$ are both true. If my understanding is correct, we are practically generalizing $A$ to the extent it embraces $B$ or vice-versa.

Choose, say, $G =$ Goldbach's conjecture, and $R =$ Riemann's conjecture. Suppose they are independent. Suppose someone generates $E = G \land R$. ($E$ would be a nice expression!) Does proving $E$ proves $G$ and proves $R$, too? It seems to be counter-intuitive for me, but - as a layman - I cannot tell you why.

  1. Are my thoughts about $C$ wrong?
  2. If not, does the proof on $E$ holds in the terms of proving both $G$ and $R$?

Bonus question: would a solution like this be shocking for the mathematical community for several (at least two) major conjectures?

  • $\begingroup$ Everything you said is true and it can't be proven per se. It is axiomatically true. From $A\land B$ you can conclude that both $A$ and $B$ are true and if both $A$ and $B$ are true, then so is $A\land B$. It wouldn't surprise anyone. $\endgroup$ – Git Gud Jul 21 '13 at 18:27
  • $\begingroup$ It is 01:00 UTC, I am going to pick an OKAY answer in 10-12 hours. :) Thank you very much all for your help! $\endgroup$ – ysfhk Jul 21 '13 at 21:00

Your thoughts about C are not wrong. By definition, A and B is true when A is true and B is true.

Now, regarding you intuitive thoughts. It is very hard to come up with an expression for A and B, other than A and B. Try it. Let A be the Pythagorean theorem, and B be the Fundamental Theorem of Arithmetic. The thing about proving A and B, is that to do it, you usually prove A first, then prove B.

Of course, there is another way. What people do often is make up an expression C, such that C implies A, and C implies B. Then they try to prove C. Mind you, this is not easy.

As another point. Let us consider Goldbach's conjecture, and how it is likely to be proved. I mean, it has been here for a long time, and people have been at it. People have made "progresses", maybe. Well, what are these progresses?

One of these progresses would the proof of something that may help prove GoldBach's conjecture. So what does that mean? Well, basically, some mathematicians guessed that D and E and some other facts together implies GoldBach's conjecture. Mind you it is a guess. And the mathematicians don't know what these other facts might be. But still, they prove D. So, if they guessed right, then they proved one part of the things needed to prove GoldBach's conjecture.

For your bonus question. The answer is yes. If A and B are two major theorems, it would be very hard to express A and B as anything other than A and B.


The rules of inference that comprise the meaning of logical and are $$\begin{align}&A\\& B\\&\overline{A\land B}\end{align} $$ and $$\begin{align}&\underline{A\land B}\\& A\end{align} $$ and $$\begin{align}&\underline{A\land B}\\& B\end{align} $$ So yes, a proof of $E$ would immediately yield a proof of $G$ and a proof of $R$.


Yes. If we can prove $A\land B$ then we can prove $A$ and we can prove $B$.

Note that by assuming that $A$ and $B$ are independent you cannot have that $C$ is provable, unless at least one of them is provable (and in particular, not independent).

  • $\begingroup$ Oh yes, Gödel successfully blows my mind with this pesky incependency-problem. $\endgroup$ – ysfhk Jul 21 '13 at 18:32


Let $x,y \in \mathbb{R}$ and $A$ denote $x = 0$ and $B$ be $y = 0$. Now, consider $C$ as

$$x^2 + y^2 = 0.$$

If you prove $C$, then you have proved $x = 0$ and $y = 0$, that is $A$ and $B$. However, proving $\neg C$, that is $x^2 + y^2 \neq 0$ proves only $\neg A \lor \neg B$, that is $x \neq 0$ or $y \neq 0$.

I doubt it would surprise anyone, but I hope it still might be helpful to you ;-)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.