0

Trying to solve a Boolean algebra expression which simplifies midway down to

$$(Q \lor (M \land \lnot N \land \lnot G)) \land (M \lor N \lor G)$$

It seems the final result of the distribution should be

$$(Q \land M) \lor (Q \land N) \lor (Q \land G) \lor (M \land \lnot N \land \lnot G)$$

However, Wolfram Alpha lists the result (shown here: https://www.wolframalpha.com/input/?i=%28Q+%7C%7C+%28~G+%26%26+M+%26%26+~N%29%29+%26%26+%28M+%7C%7C+N+%7C%7C+G%29) as

$$(Q \land N) \lor (Q \land G) \lor (M \land \lnot N \land \lnot G)$$

How did the $$Q \land M$$ term get cancelled out?

aftrumpet
  • 121
  • Did you type the initial equation correctly? At the very least, the first term needs some parentheses. – rogerl Jan 25 '16 at 20:38
  • I added parentheses to the expression in WA (besides the fact that it seems to automatically give ands parentheses), but that did not change the answer at all. – aftrumpet Jan 25 '16 at 20:41

1 Answers1

0

$Q \wedge M = (Q \wedge M \wedge N) \vee (Q \wedge M \wedge \lnot N)$ $= (Q \wedge M \wedge N) \vee (Q \wedge M \wedge \lnot N \wedge G) \vee (Q \wedge M \wedge \lnot N \wedge \lnot G)$

Now use $(Q \wedge M \wedge N) \vee (Q \wedge N) = Q \wedge N \wedge (1 \vee M) = Q \vee N$ and similar equations for the last two terms.

Of course nobody would actually try to minimize the expression in this way. There are systematic tools like Karnaugh maps or the Quine-McCluskey algorithm.

Dominik
  • 19,963