# Is This Logic Diagram Correct?

1. Apr 14, 2013

### ibcoding

The boolean expression I am working with is:

((ab)'(b'c)' + a'bpc') where ' is the NOT symbol.

I have the following logic circuit diagram. Is it correct?

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2. Apr 14, 2013

### cepheid

Staff Emeritus
Welcome to PF!

The diagram is confusing a bit, because one of the a lines is very close to one of the b lines. However it looks correct to me.

3. Apr 14, 2013

### ibcoding

Ok, I know exactly which line you are talking about; I thought the same thing. I'll change that around. Thanks!

4. Apr 14, 2013

### cepheid

Staff Emeritus
I think you can simplify this alot using DeMorgan's and other things. For instance, take the first input to the OR gate:

(ab)' * (b'c)'

Using DeMorgan's theorem (xy)' = x' + y', you can expand each thing in parentheses:

= (a' + b') * (b + c') [eq. 1]

and now using the distributive property of AND operator (which I believe is present for Boolean algebra?)

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

and using the associative property:

a'(b + c') + 0 + b'c'

Anything OR'd with 0 is just itself, so this becomes:

a'(b + c') + b'c' [eq. 2]

I'm not sure that this is any simpler in this case, but the point is that you could, in principle, try to minimize the number of gates. I wrote a little program to generate the truth tables for eq. 1 and eq. 2 and they came out the same.

Last edited: Apr 14, 2013
5. Apr 14, 2013

### ibcoding

Alright, variable p is missing from that equation, but I get the idea. On that note, is this the correct truth table outputs:

1
1
0
0
1
1
1
1
1
1
0
0
0
0
0
0

6. Apr 14, 2013

### ibcoding

Ahh, the a' a' are adjacent so it is absorbed, got ya

7. Apr 14, 2013

### ibcoding

So the equation should be a'(b + c') + b'pc'

8. Apr 14, 2013

### ibcoding

which gives me three and gates and two or gates

9. Apr 14, 2013

### cepheid

Staff Emeritus

Well, I was ONLY considering the (ab)'(b'c)' in the original equation, as an example. I wasn't trying to simplify the whole thing.

Here's what I get:

Code (Text):

a   |   b   |   c   |   p   |   f
0   |   0   |   0   |   0   |   1
0   |   0   |   0   |   1   |   1
0   |   0   |   1   |   0   |   0
0   |   0   |   1   |   1   |   0
0   |   1   |   0   |   0   |   1
0   |   1   |   0   |   1   |   1
0   |   1   |   1   |   0   |   1
0   |   1   |   1   |   1   |   1
1   |   0   |   0   |   0   |   1
1   |   0   |   0   |   1   |   1
1   |   0   |   1   |   0   |   0
1   |   0   |   1   |   1   |   0
1   |   1   |   0   |   0   |   0
1   |   1   |   0   |   1   |   0
1   |   1   |   1   |   0   |   0
1   |   1   |   1   |   1   |   0

10. Apr 14, 2013

### ibcoding

Sorry, I wasn't complaining. :) Alright, that's the same as what I have.

11. Apr 14, 2013

### ibcoding

The equation breaks down to a'b + b'c' + a'bp, I believe.