It has taken me 3 months to finally solve this problem´.

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/,
Andreas