Set Q is not computable.
Proof. Q is nontrivial because some programs satisfy the required property for membership in Q, and some do not. (Some programs fail to accept all inputs. So don't.)
Q respects equivalence. Suppose that p and q are two equivalent programs.
| p ∈ Q | ⇒ | L(p) = ∅ | |
| ⇒ | φp(x) ≠ yes for all x | ||
| ⇒ | φq(x) ≠ yes for all x | (since p ≈ q) | |
| ⇒ | L(q) = ∅ | ||
| ⇒ | q ∈ Q |
By Rice's theorem, B is uncomputable.