If you make a Karnaugh map, and put an "X" into each place where
$(B \land \lnot C) \lor (A \land B) \lor (A \land C)$ is true,
these cells are also covered by $(A \land C) \lor (B \land \lnot C)$.
| A | -A |
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B | X | | C
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B | X | X | -C
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-B | X | | C
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-B | | | -C
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To show that $(B\land \lnot C) \lor (A \land B) \lor (A \land C)$ implies
$(B\land \lnot C) \lor (A \land C)$, we can reason by cases. If the first or third disjunct holds, we are done. If $(A \land B)$ holds, then there are two subcases: if $C$ holds then $(A \land C)$ holds, and if $\lnot C$ holds then $B \land \lnot C$ holds.
This reasoning can be written using Boolean algebra laws as:
- Given $(B\land \lnot C) \lor (A \land B) \lor (A \land C)$
- because $C \lor \lnot C = 1$, and $X \land 1 = X$, is equivalent to $(B\land \lnot C) \lor [ (A \land B) \land (C \lor \lnot C)] \lor (A \land C)$
- by distributivity, is equivalent to $(B\land \lnot C) \lor [ (A \land B \land C) \lor (A \land B \land \lnot C) ] \lor (A \land C)$
- by weakening $\land$ twice, implies $(B\land \lnot C) \lor [ (A \land C) \lor (B \land \lnot C) ] \lor (A \land C)$
- is equivalent to $(B\land \lnot C) \lor (A \land C)$
You can see how using $C \lor \lnot C$ and then applying distributivity allows us to use Boolean algebra laws to simulate proof by cases.
The reverse implication, that $(B\land \lnot C) \lor (A \land B) \lor (A \land C)$ is implied by $(B\land \lnot C) \lor (A \land C)$, follows from weakening the $\lor$.