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.
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
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.