Boolean Algebra Simplification Question

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elimenohpee

Joined Oct 26, 2008
47
I have a question about a simplification in my textbook. They define a F(A,B,C) in terms of the minterms: \(\Sigma\)m(0,1,2,3,4,5). From the minterms, its easy to get an expression as the sum of products which turns out to be:

A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C

Now using a K-map and using just the boolean postulates, I simplify the expression to just:

A' + B'

But my question is, how do you compare the truth tables of these 2 functions to see if they have the same output? I don't see how just because the literal C nulls out after all the simplification.
 

Ratch

Joined Mar 20, 2007
1,070
elimenohpee,

But my question is, how do you compare the truth tables of these 2 functions to see if they have the same output? I don't see how just because the literal C nulls out after all the simplification.
You can't. Your truth table will be 8 entries containing all the minterms of a 3 variable expression. You have to expand A' + B' to get the minterms. No need to do so, however, if you did the reduction correctly.

Expanding expression to minterms:

A' + B'
A'(B +B') + B'(C + C')
A'B + A'B' + B'C + B'C'
A'B(C + C') + A'B'(C + C') + (A + A')B'C + (A + A')B'C'
A'BC + A'BC' + A'B'C + A'B'C' + AB'C + A'B'C + AB'C' + A'B'C;
A'BC +A'BC' + A'B'C + A'B'C' + AB'C + AB'C'

Which is what you started out with, and can now be made into a truth table. If you look at the truth table, you will see that variable "C" has no influence at all on the output. Everything is determined by the values of variables "A" and "B'.

Ratch
 
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