# simplifying using Boolean Algebra.

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?

• @rschwieb I edited, sorry for confusion. – Xabi May 8 '12 at 13:23
• The notation in the laws and the notation in the question are still completely different. If you think my solution below is answering the question you asked, I can attempt to make your post clear. – rschwieb May 8 '12 at 13:31
• Are the second and third terms $\overline{A}\overline{B}C$ or are they $\overline{AB}C$? – robjohn Jan 17 '13 at 21:20

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:

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

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

• This is perfect Thank you! – Xabi May 8 '12 at 13:50

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

• A bit of explanation in English words would be helpful for everyone concerned. – Robert Soupe Jul 21 '15 at 0:36

=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)