Verified answer

Your problem description is ambiguous, so I'll solve what

I *think* you intended. But feel free to update the question

with a parenthesized expression and I will be happy to

re-solve it. [Maybe shoot me an email if you update it

so that I remember to do it :-) ]

My confusion comes from you writing:

A"B'

which you describe as meaning that there is a double bar over

the A and a single bar over the B. But are these bars only

over the variables or do they extend between them. In the

first case, it would be equivalent to:

A'' & B'

in the second case, it would be equivalent to:

(A' & B)'

I expect you meant the latter, so your expression is:

f = ( (AC)'' + (A'B)' ) (B'' + C')

Although even here I'm worred that the "+" in the final

expression might be under a complement bar.

However, a double complement simply cancels out, so:

(AC)'' = AC

thus we now have:

f = ( AC + (A'B)' ) (B + C')

Apply DeMorgan's theorem to the expression (A'B)':

f = ( AC + (A + B') ) (B + C')

f = ( AC + A + B' ) (B + C')

Then group, factor, and reduce the first expression:

f = ( (AC + A) + B' ) (B + C')

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

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

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

Then distibute the AND operator (ie. "multiply"):

f = AB + B'B + AC' + B'C'

f = AB + 0 + AC' + B'C'

f = AB + AC' + B'C'

.

A'B'C' + AB'C + A'B'C A'B'C' + AB'C + A'B'C + A'B'C A'B' (C' + C) + B'C (A + A') A'B' + B'C bear in innovations you could reproduction words to decrease different words. seem at which of the unique words could be duplicated to decrease AB'C.