Propositional logic and distributive law

Question

I am having trouble trying to understand how this question passes from this point

$$ ( ( p\vee q )\wedge (p \vee \neg r ) \wedge (\neg q \vee \neg r ) ) \vee ( \neg p \vee r ) $$

to this $$ (p\vee q \vee \neg p \vee r)\wedge(p \vee \neg r \vee \neg p \vee r)\wedge(\neg q \vee \neg r \vee \neg p \vee r) $$

$$ T \wedge T \wedge T = T $$

I'm sure it has to do something with the distribution law ($p\vee(q \wedge r) =(p\vee q)\wedge(p \vee r)$) but I'm confused on how it is applied. Can anyone give me a heads-up on where and how I should start expanding?

score 3 · Accepted Answer · answered Oct 13 '14 at 22:06

3

The first formula is of the form $$(A\land B\land C)\lor D$$ while the second one is $$(A\lor D)\land(B\lor D)\land(C\lor D)\,.$$

answered Oct 13 '14 at 22:06

Berci

90,745

Got it thanks! One more question though. Applying the formula I get this $ ((p\vee q) \vee( \neg p \vee r))\wedge((p \vee \neg r )\vee(\neg p \vee r))\wedge((\neg q \vee \neg r) \vee (\neg p \vee r)) $
Does that mean I can remove the brackets and why?
– εν οίδα ότι ουδέν οίδα Oct 13 '14 at 22:20
2

Nevermind associative law sais yes. Thanks – εν οίδα ότι ουδέν οίδα Oct 13 '14 at 22:39

Propositional logic and distributive law

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