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Can you please check this boolean algebra answer?
Z = [A\ (B + C D\ ) + (AC\)\ ]\
\ means 'bar'.
i.e. Z is equal to (A-bar (B + C D-bar)+ (A C-bar)-bar) the whole bar
My solution:
Z = (A\ (B+CD\) + (AC\)\ )\
= (A\ (B+CD\))\ . (AC\)\\
= (A\\ + (B+CD\)\ ) . (AC\)
= (A + (B\.(CD\)\) ) . (AC\)
= (A + (B\. (C\+D)) . (AC\)
= (A + B\C\ + B\D) . (AC\)
= AC\ + AB\C\ + AB\C\D
= AC\ (1 + B\ + B\D)
= AC\ <==
1 Answer
- morganLv 71 decade agoFavorite Answer
To check your work, I used a different derivation:
z = (a'(b+cd')+(ac')')'
z = (a'b + a'cd' + a' + c)'
z = (a'(b + cd' + 1) + c)'
z = (a'(1) + c)'
z = (a' + c)'
z = ac'
Then to *double* check, I entered the equation into
the solver at http://www.wolframalpha.com/ using their
syntax:
~(~a &&(b||(c&&~d))||~(a&&~c))
and again verified that the reduced expression is:
z = ac'
I mention this because you might find it useful to
use that same solver to check your work.
.
