In Sipser's book of Theory of Computation, related to Reducibility, it's written

if A is undecidable and reducible to B, B is undecidable.

The confusion is, only a solution to B determines a solution to A, if i'm not wrong. So, for instance if B is decidable and A is undecidable, what does it mean? Here B isn't undecidable.

Hope you got it.


Suppose $A$ is undecidable and reducible to $B$. Given an algorithm for $B$, you can use the reduction from $A$ to $B$ to get an algorithm for $A$, which we assumed is impossible. Therefore $B$ must be undecidable.

If $A$ is undecidable and reducible to $B$ and $B$ is decidable then we obtain a contradiction and everything goes. Pigs can fly.

| cite | improve this answer | |

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.