Reduction of Boolean Expression to its Lowest From

George SA
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Hey Guys

I am Currently doing my degree in information systems. At the moment the subject is PLC(Processing and Logic Concepts)

I understand all of it, I am not sertain on how to start reducing a boolean expression. Once the first step is done I normally get it and can complete the process of reduction using the Laws.

Does anyone know if i hint or a trick on where to start reducing a Boolean Expression

EX. F = AB'C + ABC + A'B'C' + A'BC'

I am not looking for the answer to the expression only a tip on how to start reducing these tipes of Expressions.

Thank You in Advance
 
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I'm no expert on this but the first thing I would do is start looking for "similarities" in the terms. For example, I see both A and C in the first two, both A' and C' in the second two:

F= AB'C+ ABC+ A'B'C'+ A'BC'= A(B+ B')C+ A'(B+ B')C'
and, of course, those two now have the same "B+ B' " so is the same as
F= (B+ B')(AC+ A'C')_

You haven't said what you consider "reduced to its lowest form".
 
HallsoIvy

Thank You for your reply. It helped me alot. What I mean by Reduced to its lowest form is - The Boolean Expression is reduced to a point where no more Laws or Rules can be applied to the remainder of the original expression.

Would I be correct if I reduced the expression from where you stopped to :

F = (B+B')(AC+A'C')
= (1)(AC+A'C')
= AC + A'C'
= AC

Thank You for your help
 
George SA said:
HallsoIvy

Thank You for your reply. It helped me alot. What I mean by Reduced to its lowest form is - The Boolean Expression is reduced to a point where no more Laws or Rules can be applied to the remainder of the original expression.

Would I be correct if I reduced the expression from where you stopped to :

F = (B+B')(AC+A'C')
= (1)(AC+A'C')
= AC + A'C'
= AC

Thank You for your help

Stop at F = AC + A'C', you can't justify the last line.
 
Thanks I see what you mean. I went to class last night and the same was said to me on how to start these expressions. Look for similarities, Thank You again for your assistance I am almost sure I got this now
 
The last line can actually be done using an XNOR gate.

BiP
 
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