Simplifying a 4-term equation using boolean algebra

Question

1432928

So, I have a logical expression:

$(\lnot A\land B\land C\land \lnot D)\lor (\lnot A\land B\land C\land D)\lor (A\land \lnot B\land \lnot C\land D)\lor (A\land \lnot B\land C\land D)\lor (A\land B\land \lnot C\land D)\lor (A\land B\land C\land \lnot D)$ .

Part of the reason I'm having trouble is due to the length of the expression.

So far, I've managed to use complement and identity laws to cancel out one pair of Cs and Ds, and ended up with this:

$(\lnot A\land B\land C)\lor (A\land \lnot B\land D)\lor (A\land B\land \lnot C\land D)\lor (A\land B\land C\land \lnot D)$

Now, I know, through using Karnaugh maps, that the answer is this:

$(\lnot A\land B\land C)\lor (B\land C\land \lnot D)\lor (A\land \lnot C\land D)\lor (A\land \lnot B\land D)$ .

So, I'm close. I have two of the parameters. But I need to simplify two others.

I'm not asking for the straight answer, but can I have a hint as to what I need to do next?

Answer 1

$\lnot A\land B\land C\land\lnot D\quad$ can be combined with
$\lnot A\land B\land C\land D\quad$ to give
$\lnot A\land B\land C\quad$ .

$\lnot A\land B\land C\land\lnot D\quad$ can be combined with
$A\land B\land C\land\lnot D\quad$ to give
$B\land C\land\lnot D\quad$ .

$A\land\lnot B\land\lnot C\land D\quad$ can be combined with
$A\land\lnot B\land C\land D\quad$ to give
$A\land\lnot B\land D\quad$ .

$A\land\lnot B\land\lnot C\land D\quad$ can be combined with
$A\land B\land\lnot C\land D\quad$ to give
$A\land\lnot C\land D\quad$.

$\hline$

$\lnot A\land B\land C\quad$ can be combined with
$B\land C\land\lnot D\quad$ to give
$B\land C\land\lnot(A\land D)$
$(\lnot A\lor\lnot D)\equiv \lnot(A\land D)\quad$.

$A\land\lnot B\land D\quad$ can be combined with
$A\land\lnot C\land D\quad$ to give
$A\land D\land\lnot(B\land C)$
$(\lnot B\lor\lnot C)\equiv\lnot(B\land C)\quad$.

$\hline$

So, $(B\land C\land\lnot(A\land D))\lor(A\land D\land\lnot(B\land C))\quad$.

$\hline$

$(A\land D)\not\equiv(B\land C)$