So I've got an expression I have been trying to simplify and have the answer but I can't figure out how to get to it... can anyone help me out?
Equation: $(A\wedge \lnot B \wedge \lnot C \wedge \lnot D) \vee (C \wedge \lnot D) \vee (C \wedge \lnot D) = (A \wedge \lnot B \wedge \lnot D) \vee (C \wedge \lnot D)$
A Karnaugh map can help. You have already invoked Idempotent Law to eliminate the repeated term $(C \wedge \lnot D)$. Let $T$ represent a tautology (always true). Then to simplify further, observe that: $$ \begin{align*} (C \wedge \lnot D) &= (T) \land (C \wedge \lnot D) & \text{by Identity Law}\\ &= [(A \land \neg B) \lor T] \land (C \wedge \lnot D) & \text{by Domination Law}\\ &= [(A \land \neg B) \land (C \wedge \lnot D)] \lor [T \land (C \wedge \lnot D)] & \text{by Distributive Law}\\ &= (A \land \neg B \land C \wedge \lnot D) \lor (C \wedge \lnot D) & \text{by Identity Law} \\ \end{align*}$$
Basically, we are able to introduce an extra term for free that will help us factor the original expression. Returning to the original problem, we obtain:
$$ \begin{align*} &(A\wedge \lnot B \wedge \lnot C \wedge \lnot D) \vee (C \wedge \lnot D) \\ &= (A\wedge \lnot B \wedge \lnot C \wedge \lnot D) \vee (A \land \neg B \land C \wedge \lnot D) \lor (C \wedge \lnot D) & \text{by substituting above}\\ &= [(A\wedge \lnot B \wedge \lnot D) \land (\neg C \lor C)] \lor (C \wedge \lnot D) & \text{by Distributive Law}\\ &= [(A\wedge \lnot B \wedge \lnot D) \land (T)] \lor (C \wedge \lnot D) & \text{by Inverse Law}\\ &= (A \wedge \lnot B \wedge \lnot D) \vee (C \wedge \lnot D) & \text{by Identity Law} \\ \end{align*}$$ as desired.