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I had the following boolean expression:
$$(C\lor D)\land({\sim} B\lor D)\land({\sim} A\lor {\sim} C\lor {\sim} D)$$

I know this can be simplified to:
$$(C\land{\sim}B\land{\sim} D)\lor(D\land{\sim} A)\lor(D\land{\sim} C)$$ I have gotten this far:
$$({\sim} D\land C\land{\sim} B)\lor(C\land{\sim} B\land{\sim} A)\lor(D\land{\sim} A)\lor(D\land{\sim} C)$$ I have tried and tried, but still can't figure out how to get rid of the extra $$(C\land{\sim} B\land {\sim} A)$$

Please help. Thank you!

Edit: Bram28 pointed out that I mistyped, but I still cant find a way to get rid of the extra $(C\land{\sim} B\land {\sim} A)$

egreg
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TheAmazingChuck
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1 Answers1

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You can get rid of the extra

$$(C\land{\sim} B\land{\sim} A)$$

term by expanding it to

$$(C\land{\sim} B\land{\sim} A \land D)\lor (C\land{\sim} B\land{\sim} A \land \sim D)$$

Then, the first of these gets absorbed by $(D \land \sim A)$, while the second gets absorbed by $(C\land{\sim} B \land \sim D)$

Bram28
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  • Yes you are right, I mixed them up, but fixed it now, I still can't understand how to lose the (C∧∼B∧∼A). – TheAmazingChuck Dec 10 '18 at 19:13
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    @TheAmazingChuck Expand $C \land \sim B \sim A$ to $(C \land \sim B \land \sim A \land D) \lor (C \land \sim B \land \sim A \land \sim D)$, and see if each of them can get absorbed. You know Absorption right? – Bram28 Dec 10 '18 at 20:41
  • Yes I got it thank you! – TheAmazingChuck Dec 10 '18 at 22:08
  • @ryang Not sure why you thought I'd be an expert on probability theory .. but sure, I answered that question to the best of my ability. :P – Bram28 Mar 23 '23 at 02:31