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.

 

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)

 

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.

 

Just multiply the terms most of them will turn out to be zero.
Then use simple boolean logic.
The solution is here : http://www.wolframalpha.com/input/?i=%28a+or+b+or+not+c%29+and+%28not+a+and+not+b+or+c%29

 

Thanks guys, I appreciate the help.

I actually used a karnaugh map to solve it as that is way easier for me, but the HW required just striaght algebraic manipulation.