Reducibility in Computability Theory
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.