simplifying the boolean expression (a'b'c) + ('a'bc) + ('ab'c) + (a'bc)

Question

I'm having a lot of trouble on the last part of a boolean expression I'm trying to simplify. so far I got (where multiplication is AND and addition is OR):

(a * 'b * 'c) + ('a * 'b * c) + ('a * b *'c) + (a * 'b * c)

(a * 'b * 'c) + (a * 'b * c ) + ('a * 'b * c) + ('a * b * 'c)

a(('b * 'c) + ('b * c)) + ('a * 'b * c) + ('a * b * 'c)

a('b(c + 'c)) + ('a * 'b * c) + ('a * b * 'c)

a('b(1)) + ('a * 'b * c) + ('a * b * 'c)

(a * 'b) + ('a * 'b * c) + ('a * b * 'c)

however, the answer i got from wolfram alpha is

(a * 'b) + ('b * c) + ('a * b * 'c)

i just have no idea how to get the last step done. Any help would be appreciated

If `*` is AND what's `bc`, for example? I think you're having troubles with the fact that *s are interpreted as markup...and which is negation 'x or x'? — ChiefTwoPencils, Apr 07 '15 at 23:34
Hint: Using the fact that `x = x + x * y`, expand `(a * 'b)` to `(a * 'b) + (a * 'b * c)`. — Michael Liu, Apr 07 '15 at 23:53
[Wikipedia](http://en.wikipedia.org/wiki/Boolean_algebra#Monotone_laws) calls this "absorption". — Michael Liu, Apr 13 '15 at 02:27

score 0 · Answer 1 · answered Aug 29 '17 at 07:50

Your four terms (any of which make the entire expression true) are:

1/ a  b' c'
2/ a' b' c
3/ a' b  c'
4/ a  b' c

What we're trying to do is to eliminate irrelevant items so we look for pairs where two items are identical.

To get Alpha's result, first combine 1 and 4. The commonality there is ab' so whether c is true or false is irrelevant. And, since we've combined them, neither 1 nor 2 is necessary for the expression any more.

Then combine 2 and 4. The commonality there is b'c so a has no effect. Again, combining these two means that they are no longer needed in the final expression.

So 1, 2, and 4 are gone. There is no two-item commonality between 3 and any other sub-expression so there's no further simplification possible.

This gives us:

a^b'  v  b'^c  v  a'^b^c'

simplifying the boolean expression (a*'b*'c) + ('a*'b*c) + ('a*b*'c) + (a*'b*c)

1 Answers1

simplifying the boolean expression (a'b'c) + ('a'bc) + ('ab'c) + (a'bc)