[Boolean algebra] Mismatch between algebra and karnaugh map?

Josh Samman

Joined Mar 29, 2013
11
Look at this equation:

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

I solve it by distributing the terms:

f = a'(a + c) + b(a + c)
= aa' + a'c + ab + bc
= a'c + ab + bc --> Resolution

Then i plot it into the Karnaugh map:

* Sorry for the bad drawing

And i get:

f = a'c + ab(from the map)

f
= a'c + ab + bc --> (comparing to the resolution by algebra the "bc" term is missing).

I look at the map and i know right away that the "bc" term gets simplified by ther other 2 (regarding that the bold 1's in the map are the "bc" term).

Now i don't know how i can simplify f = a'c + ab + bc to f = a'c + ab using algebra only! How to do it? There's any rule i'm missing in there or what?

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MrChips

Joined Oct 2, 2009
22,069
Expand each of the three terms a'c + ab + bc

For example,

a'c = a'b'c + a'bc

Take it from there.

Josh Samman

Joined Mar 29, 2013
11
Expand each of the three terms a'c + ab + bc

For example,

a'c = a'b'c + a'bc

Take it from there.
F = (a'bc + a'b'c) + (abc + abc') + (abc + a'bc)
= a'bc+ a'b'c + abc + abc';
= [a'c(b+b')] + [ab(c'+c)]
= a'c + ab

\o/ - Thank you man, for the help