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: $(A\wedge \lnot B \wedge \lnot C \wedge \lnot D) \vee (C \wedge \lnot D) \vee (C \wedge \lnot D) = (A \wedge \lnot B \wedge \lnot D) \vee (C \wedge \lnot D)$

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    $\begingroup$ Compare the terms on the left and right. Try to only move what is absolutely necessary to make what's on the left look like what's on the right, using the distributive laws for $\vee$ and $\wedge$. The identity $X \vee X = X$ might also be useful. $\endgroup$ – Zach L. Jun 9 '13 at 5:23
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    $\begingroup$ Hmmm see what I'm really having trouble with is seeing where ¬C went to in A∧¬B∧¬C∧¬D... I've got the equation simplified to (A∧¬B∧¬C∧¬D)∨(C∧¬D) and i'm not sure if there is a rule I don't know yet or if its just blindingly simple and I don't realize it $\endgroup$ – Eric Jun 9 '13 at 5:30
  • $\begingroup$ So you've noticed that you need to get $\lnot C$ out of that big term. What does the distributive law say if you want to pull something out of a big chain of $\wedge$'s with a $\vee$ next to it, and then how can you combine that with $(C \wedge \lnot D)$ once you get it out? $\endgroup$ – Zach L. Jun 9 '13 at 5:34

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.


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