Proof writing involving Boolean algebra: AB' + AC + BC

Question

How?

AB' + AC + BC ≡ AB' + BC

RS

≡ AB' + AC + BC

≡(AB' + A)(C + BC)

≡ AC

Am I missing something? Thanks.

score 2 · Accepted Answer · answered Nov 19 '15 at 02:30

2

A quick test with truth values would show that what you have cannot be right: if $A$ is false and $C$ is true, $AC$ is false, but $AB'+BC$ is true when $B$ is true.

$$\begin{align*} AB'+AC+BC&\equiv AB'+A(B+B')C+BC\\ &\equiv AB'+ABC+AB'C+BC\\ &\equiv(AB'+AB'C)+(ABC+BC)\\ &\equiv AB'+BC \end{align*}$$

answered Nov 19 '15 at 02:30

Brian M. Scott

631,399

OMG im so dumb! – Mathy Nov 19 '15 at 02:32
Can this be translated to operation "Sets" in the similar way? – Mathy Nov 19 '15 at 02:34
2

@Steve: Don’t worry: we’ve all felt that way on occasion! \ You mean with intersection, union, and complement? Yes. – Brian M. Scott Nov 19 '15 at 02:35
Just to make sure, (A∩B')∪(A∩(B∪B′)∩C)∪(B∩C), this statement is valid and will yield (A∩B′)∪(B∩C) as well right? – Mathy Nov 19 '15 at 02:42
@Steve: Yes, using essentially the same sequence of steps. – Brian M. Scott Nov 19 '15 at 02:43
Okay, thanks so much :) have a really good night! – Mathy Nov 19 '15 at 02:44
@Steve: You’re welcome; you too! – Brian M. Scott Nov 19 '15 at 02:45

Proof writing involving Boolean algebra: AB' + AC + BC

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