# Homework - Boolean Simplification

Discussion in 'Homework Help' started by SaadMo, Sep 15, 2014.

1. ### SaadMo Thread Starter New Member

Sep 15, 2014
2
1
Hi guys, I am new to the forums but have benefited from the sites tutorials.

I have a hw problem: To simplify (a+b+c')(a'b'+c)

If someone could work it out step by step, how they would do it, I would really appreciate it. What I typically do is take the dual of the function, then start working on the a'+b'c side by using x+x' to make sure all the terms have the variables.

I end up with a'bc+a'b'c'+ab'c+a'b'c, but I am not sure how to proceed from here, or even if I did the earlier steps correctly.

2. ### JoyAm Member

Aug 21, 2014
112
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Have you considered using a Karnaugh map ? I believe its the best way to solve stuff like that.
I got a'c+ab'+b'c ( i havent worked on it lately so i may be wrong)

3. ### MrChips Moderator

Oct 2, 2009
12,636
3,455
To simplify (a+b+c')(a'b'+c)

multply the terms (a'b'+c) by a, then by b and by c'

and then simplify the SOP (sums of products).

Yes, Karnaugh mapping is a very useful tool for simplifying boolean logic. However, it does not help to reduce the expression in this example.

Aug 9, 2014
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