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The expression I need help simplyfing is:

$\lnot$ A$\lnot$B$\lnot$C$\lnot$D + A$\lnot$B$\lnot$C + A$\lnot$BC$\lnot$D + ABD + $\lnot$A$\lnot$BC$\lnot$D + B$\lnot$CD + $\lnot$A

which can be simplified into:

$\lnot$A + BD + $\lnot$B$\lnot$C + $\lnot$B$\lnot$D

So far I have come up with:

$\lnot$A($\lnot$B$\lnot$C$\lnot$D + $\lnot$BC$\lnot$D + 1) + A$\lnot$B$\lnot$C + A$\lnot$BC$\lnot$D + ABD + B$\lnot$CD =

$\lnot$A + $\lnot$AB($\lnot$C + C$\lnot$D) + ABD + B$\lnot$CD =
$\lnot$A + A$\lnot$B$\lnot$C + A$\lnot$B$\lnot$D + ABD + B$\lnot$CD

This is where I get stuck and can't see how I can work backwards from the solution to get to where I am now.

usernamn
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1 Answers1

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Let me use the complement notation in place of the negation one.
Let us also start to denote by $F(A,B,C,D)$ the expression $$A'B'C'D'+AB'C'+AB'CD'+ABD+A'B'CD'+BC'D+A'.$$

Notice that by absorption, $A'$ eliminates all the expressions with $A'$, so that $$F(A,B,C,D)=AB'C'+AB'CD'+ABD++BC'D+A'.$$

Using the distributive law, we get $$F(A,B,C,D)=AB'(C'+CD')+ABD+BC'D+A'.$$

Now we use the equation

$$(X'+XY)=(X'+X)(X'+Y)=X'+Y.$$

In particular, $C'+CD'=C'+D'$, and we obtain $$F(A,B,C,D)=AB'C'+AB'D'+ABD+BC'D+A'.$$

Using the distributivity and commutativity, $$F(A,B,C,D)=A'+A(BD+B'C'+B'D')+BC'D,$$ and using the highlighted equation once more, $$F(A,B,C,D)=A'+BD+B'C'+B'D'+BC'D.$$ By absorption, $BD+BC'D=BD$, yielding the desired result.

amrsa
  • 12,917
  • I see, so you're treating $(BD+B'C'+B'D')$ as a single term to use the equation and eliminate $A$?. This was really helpful, thanks! – usernamn Nov 13 '19 at 21:03
  • @usernamn Yes, that's it. It applies to any term. So replace $Y$ with that one. – amrsa Nov 14 '19 at 10:58