Is This Boolean Expression Simplified Correctly?

[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?

 

[cepheid]

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.

 

[ibcoding]

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

 

[cepheid]

I think you can simplify this a lot 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.

 

[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

 

[ibcoding]

I think you can simplify this a lot 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.

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

 

[ibcoding]

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

 

[ibcoding]

which gives me three and gates and two or gates

 

[cepheid]

Alright, variable p is missing from that equation, but I get the idea.

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.

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

Here's what I get:

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

 

[ibcoding]

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

 

[ibcoding]

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