It has taken me 3 months to finally solve this problem—well, actually this is the first time in three months that I have taken a look at the problem again.
Anyway—I proudly present:
\[ (b \Rightarrow B_0) \land (\lnot b \Rightarrow B_1) \equiv \] \[ \equiv (\lnot b \lor B_0) \land (b \lor B_1) \equiv \] \[ \equiv (b \land B_0) \lor (\lnot b \land B_1) \lor (B_0 \land B_1) \equiv \] \[ \equiv (b \land B_0) \lor (\lnot b \land B_1) \lor ((b \lor \lnot b) \land B_0 \land B_1)\equiv \] \[ \equiv (b \land B_0) \lor (\lnot b \land B_1) \lor (b \land B_0 \land B_1) \lor (\lnot b \land B_0 \land B_1) \equiv \] \[ \equiv (b \land B_0) \lor (b \land B_0 \land B_1) \lor (\lnot b \land B_1) \lor (\lnot b \land B_0 \land B_1) \equiv \] \[ \equiv (b \land (B_0 \lor (B_0 \land B_1)) \lor (\lnot b \land (B_1 \lor (B_0 \land B_1)) \equiv \] \[ \equiv (b \land B_0) \lor (\lnot b \land B_1) \]
\o/