Question

i have the following problem boolean simplfication

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

I can do it two different ways factor AB' out of the 3rd and 4rth terms giving me

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

then using the boolean rules that says C'+ C =1

and the rule that says anything anded with 1 is it's self i have

A'B'C' + BC + AB' comuntive law

AB' + A'B'C + BC

now factoring B' out of the first two terms

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

now using the rule A+A'C = A +C

i Have B' (A+C') + BC

now distrubute back in B'

AB' +B'C which is wrong when I Test it...

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

the correct way to factor B'C' out of the 3rd and 1st terms giving me

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

then using the boolean rules that says A'+ A =1

and the rule that says anything anded with 1 is it's self i have

B'C' + BC + AB'C

now I factor B out of the 2nd and 3rd term

giving me B'C'+ B(C+C'A) because of the communitve law and the the boolean law

A+A'B = A +B

I have B'C'+ B(C+ A)

distubute the B back in I get

B'C'+ BC +AC which is right when i tested it..

The only difference between the two approaches is the first way i factor the common term out of the 3rd and 4th terms

and second way i factored the coman term out of 1st and 3rd terms which also have a common factor ..... so my question is how can the boolean algibra be right in both cases but lead ro two differet results ?

the first answer is AB'+ B'C'+ BC

which is worng.

Answer 1

Note- I am totally confused in question you have given A'BC' and in midway A'B'C'. Which is correct? Don't know. I used k map and simplification using Boolean algebra, and getting same for questions at the top. Please find me in comment box if you find any difficulty. Good luck.