0
$\begingroup$

I was doing the following question. Using the following rules of boolean algebra:

         _ law 1: X+X=1   law 2: X.1=X   law 3:X.Y+X.Z = X.(Y+Z) 

simplify:

    __  _   __  ABC+ABC+ABC+ABC 

I have tried to simplify starting off with law 1, but I get stuck because the complements somewhat confuse me.

The bar meaning complement or 'not'

Could someone explain?

  • 0
    Are the second and third terms $\overline{A}\overline{B}C$ or are they $\overline{AB}C$?2013-01-17

3 Answers 3

1

If I understand correctly, the problem is to reduce $ (\overline{A \& B} \& C) + (\overline{A} \& B \& C)+ (\overline{A \& B} \& C) +(A\& B \& C) $

You can deduce from the laws that $X+X=X$, so this is clearly already

$ =(\overline{A \& B} \& C) + (\overline{A} \& B \& C)+(A\& B \& C) $ By law 3, then law 1, then law 2: $ (\overline{A} \& B \& C)+(A\& B \& C)=(\overline{A}+A)\&B\& C=1\&B\& C=B\&C $, so the original expression is now:

$ =(\overline{A \& B} \& C)+B\&C $

By De Morgan's laws $\overline{A \& B}=(\overline{A}+\overline{B})$, and you can deduce the rest from your laws:

$ =(\overline{A} + \overline{B}) \& C+B \& C =(\overline{A} + \overline{B}+B) \& C $

$ =(\overline{A} + 1) \& C=1\&C=C $

  • 0
    This is perfect Th$a$nk you!2012-05-08
1

ABC+A'B'C+A'BC+A'B'C

=ABC+A'B'C+A'BC

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

=ABC+A'C+B'C+A'BC

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

=BC+AC+B'C

=C(B+B')+AC

=C+AC

=C(1+A)

=C

  • 1
    A bit of explanation in English words would be helpful for everyone concerned.2015-07-21
0

=C(AB+A'B')+ A'C(B+B') =C(1) + A'C(1) =C + A'C =C(1+A') =C according to (1+x=1)