Set B is not computable.
Proof. B is nontrivial because some programs satisfy the required property for membership in B, and some do not.
B respects equivalence. Suppose that p and q are two equivalent programs.
| p ∈ B | ⇒ | φp(1)↓ and φp(2)↓ and φp(1) = φp(2) | |
| ⇒ | φq(1)↓ and φq(2)↓ and φq(1) = φq(2) | (since p ≈ q) | |
| ⇒ | q ∈ B |
By Rice's theorem, B is uncomputable.