## Boolean expression simplification

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.
imu_1
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Joined: Wed Sep 07, 2011 2:55 pm
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### Boolean expression simplification

Hello, I am trying to simplify this expression

AC' + AB' + A'BC' + A'B'C TO AC' + BC' + B'C

This is what I have done so far:

AC' + A'BC' + AB' + A'B'C
= C'(A+A'B) + B'(A+A'C)
= C'(A+B) + B'(A+C)
= AC' + BC' + AB' + BC'

I wanna get rid off AB' but am not sure how?? Any helo will be appreciated.

les
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Joined: Fri Dec 02, 2011 4:33 pm
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### Re: Boolean expression simplification

Your question "how to eliminate the AB' expression" still stands, despite an error in the last line of your final expression.

It should be:
AC' + BC' + AB' + B'C

Despite this, drawing up the truth tables, I verified that AB' can indeed be eliminated - its value is never true when the rest of the expression is false.

Have you figured this one out? It's been a long time since I did this in school.

les
Posts: 2
Joined: Fri Dec 02, 2011 4:33 pm
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### Re: Boolean expression simplification

Got it -- here's how:

AC' + BC' + AB' + B'C <-- the expression you wanted to remove the AB' from
= AC' + BC' + AB'(C' + C) + B'C <-- AND'ing AB' with a true expression (C' + C) does not change the value of the result
= AC' + BC' + AB'C' + AB'C + B'C <-- distributing the AB'
= AC' + AB'C + BC' + AB'C + B'C <-- rearranging the terms (associative)
= AC'(1 + B') + BC' + B'C(A + 1) <-- 1 + anything = 1, anything + 1 = 1
= AC' + BC' + B'C

Eliminating the AC' seems just as valid:
AC' + BC' + AB' + B'C <-- given
=AC'(B + B') + BC' + AB' + B'C <-- same ANDing trick as above
=AC'B + AC'B' + BC' + AB' + B'C <-- rearranging
=AB' (1 + C') + BC'(A + 1) + B'C <-- 1+anything = 1, anything + 1 = 1
= AB' + BC' + B'C

I wrote out the truth table for both expressions to verify that these provided the same result.

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