How is that:
(!B AND D) OR (!C AND B) OR (!C AND D)
Equal to that:
(B AND !C) OR (!B AND D)
For the life of me I cannot figure it out, been trying for a good few hours.
The original problem was to convert this minimized product of sums:
(!B OR !C) AND (B OR D)
To sum of products (not minimized).
Use Adjacency: $(P \land Q) \lor (P \land \neg Q)\Leftrightarrow P$
$(\neg B \land D) \lor (\neg C \land B) \lor (\neg C \land D) \Leftrightarrow$ (Adjacency x 3)
$[(\neg B \land D \land C) \lor (\neg B \land D \land \neg C)] \lor$ $ [(\neg C \land B \land D) \lor (\neg C \land B \land \neg D)] \lor$ $ [(\neg C \land D \land B) \lor (\neg C \land D \land \neg B)] \Leftrightarrow$ (Association, Commutation (i.e. reorder))
$ [(\neg C \land B \land D) \lor (\neg C \land D \land B) \lor (\neg C \land B \land \neg D)] \lor$ $ [(\neg B \land D \land C) \lor (\neg B \land D \land \neg C) \lor (\neg C \land D \land \neg B)] \Leftrightarrow$ (More reordering)
$ [(B \land \neg C \land D) \lor (B \land \neg C \land D) \lor (B \land \neg C \land \neg D)] \lor$ $ [(\neg B \land D \land C) \lor (\neg B \land D \land \neg C) \lor (\neg B \land D \land \neg C)] \Leftrightarrow$ (Idempotence ($P \lor P \Leftrightarrow P$)
$ [(B \land \neg C \land D) \lor (B \land \neg C \land \neg D)] \lor$ $ [(\neg B \land D \land C) \lor (\neg B \land D \land \neg C)] \Leftrightarrow$ (Adjacency x 2)
$ (B \land \neg C) \lor (\neg B \land D)$