Reduce the unsimplified Boolean expression to it simplest form

Thread Starter

Dercha

Joined May 26, 2021
2
Hi,
I have gotten this expression from truth table.
Y=A’B’C’ + A’B’C + A’BC + AB’C + ABC

and I gotten the simplistic form with k map as Y = C + A’B’

but when I try using Boolean, I just can’t get it. I am bad at algebra so someone please advice me where i am wrong and how and why.

Y=A’B’C’ + A’B’C + A’BC + AB’C + ABC
Y = A’B’(C’+C) + BC(A’+A) + AB’C
Y = A’B’ + BC + AB’C
Y = A’B’ + C(B+AB’)
Y = A’B’ + C(B+A)
Y = A’B’ + CB + AC

can’ seems to simply anymore.
Where am I wrong?
 

Thread Starter

Dercha

Joined May 26, 2021
2
Hi D,
Welcome to AAC
How did you get from here Y = A’B’ + C(B+AB’) to here. Y = A’B’ + C(B+A)

E
Hi Eric

Thanks for reply
Y = A’B’(C’+C) + BC(A’+A) + AB’C
Y = A’B’+ BC + AB’C ( cause C’+C and A’+A) =1.
Then from BC + AB’C I took out the common factor C, and thus become C( B+AB’).
And with B + AB’ the singular B+A.
 

BobaMosfet

Joined Jul 1, 2009
1,768
Hi,
I have gotten this expression from truth table.
Y=A’B’C’ + A’B’C + A’BC + AB’C + ABC

and I gotten the simplistic form with k map as Y = C + A’B’

but when I try using Boolean, I just can’t get it. I am bad at algebra so someone please advice me where i am wrong and how and why.

Y=A’B’C’ + A’B’C + A’BC + AB’C + ABC
Y = A’B’(C’+C) + BC(A’+A) + AB’C
Y = A’B’ + BC + AB’C
Y = A’B’ + C(B+AB’)
Y = A’B’ + C(B+A)
Y = A’B’ + CB + AC

can’ seems to simply anymore.
Where am I wrong?
This is called, Boolean Algebra- very handy. maybe this tutorial will help:

https://www.tutorialspoint.com/computer_logical_organization/boolean_function.htm
 
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