# Boolean Algebra help

Discussion in 'Homework Help' started by w1zard, Jan 24, 2012.

1. ### w1zard Thread Starter New Member

Aug 5, 2010
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0
I'm struggling to work out a problem I know the answer to. I've got this far:

A'.B'.C' + B

...but I need to further reduce it to

A'.C' + B

What rules should I be applying to make that final step?

Thanks

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
You can do a Karnaugh map simplification for a proven answer.

But think about it this way. If B=1, then the expression is true anyway.
If B=0, then B'=1 and the expression depends only on A'C'.

Does that help?

3. ### djsfantasi AAC Fanatic!

Apr 11, 2010
2,905
879
I can't remember the rule name, but if you draw out a quick truth table for the remaining terms, it should be obvious. This is something from experience that I pick out of an expression.

Let Z=A'.C', then your last expression is B'.Z + B. If B=1 then the B'Z term drops out. If B=0, then Z is the result. Hence, the B' is redundant in the first term.

Last edited: Jan 24, 2012
4. ### w1zard Thread Starter New Member

Aug 5, 2010
11
0
Thanks so much for both your answers. This has made it much clearer in my head, I can see exactly why it works now.

If anyone else knows the 'proper' name of the rule that I am applying to arrive at the answer, that would also be helpful (so I can document my work) - but I can see why it works now, which I couldn't quite see before

5. ### MrChips Moderator

Oct 2, 2009
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Rather than learning the rules, learn to use a Karnaugh map. It works every time.

6. ### w1zard Thread Starter New Member

Aug 5, 2010
11
0
Unfortunately I've got to prove this works using Boolean Algebra only. I'm not allowed to use Karnaugh maps for this particular problem.

That's why I know the answer (having used Karnaugh to arrive there), but was struggling to see how to get to the same place with Boolean algebra alone.

7. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
In that case, work backwards:

A'C'+B=
A'C'(B+B')+B=
A'C'B+A'C'B'+B=
B(1+A'C')+A'C'B'

8. ### MrChips Moderator

Oct 2, 2009
12,636
3,454
So, you used the Karnaugh map incorrectly. You need to wrap the cells around as in a cylinder.

Wrapping A'B'C' with A'BC' reduces to A'C'.

9. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Actually he said that he got the answer through a Karnaugh map and it was correct.
But K-maps don't reveal you the progression of Boolean operations to reach the result, which was his problem to begin with.