How come

$$(\neg p \land q) \lor ( \neg p \land \neg q) \Leftrightarrow \neg p \land (q \lor \neg q)$$

by distributive law? I simply don't understand how they made them equivalent by the ditributive law. Could smb explain in details? Thanks!

  • $\begingroup$ Do you know what the distributive law is? $\endgroup$ – David Sep 17 '15 at 6:27
  • $\begingroup$ @David, I know but how does it even have any similarities with simple algebra? Could you please explain? $\endgroup$ – John Sep 17 '15 at 6:29
  • $\begingroup$ Please post the distributive law for logic in your question. $\endgroup$ – David Sep 17 '15 at 6:33
  • $\begingroup$ @David, I've posted it in the question $\endgroup$ – John Sep 17 '15 at 6:36
  • $\begingroup$ OK, that's correct. Now you can substitute any expressions you like for $p,q,r$. Can you see what substitutions to make so that the LHS of your distributive law is identical with the RHS of your question? And then what will you have on the RHS of your distributive law? $\endgroup$ – David Sep 17 '15 at 6:41

EDIT: There are two Distributive laws: $$(p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)$$ and $$(p \land (q \lor r) \equiv (p \land q) \lor (p \land r).$$

So we'll have $$(\neg p \land q) \lor ( \neg p \land \neg q)$$

$$\Leftrightarrow (\neg p \lor \neg p) \land ( q \lor \neg q)\text{ (By Distributivite Law)}$$

$$\Leftrightarrow \neg p \land (q \lor \neg q) \text{ (By Idempotent Law)} $$

  • $\begingroup$ Do I always have to multiply first coefficients with each other, second coefs with each other, and etc, when there is v, @OGC? $\endgroup$ – John Sep 17 '15 at 6:30
  • $\begingroup$ @John I've made an edit, and they are propositions, not coefficients. $\endgroup$ – OGC Sep 17 '15 at 6:57

Here is distributive law:- $A\wedge (B \vee C) \equiv (A \wedge B) \vee (A \wedge C)$

Start with right hand side you can understand it..
$\neg{p} \wedge (q \vee \neg{q}) \equiv (\neg{p}\wedge q)\vee(\neg{p}\wedge q)$
In This equation predicate $\neg p$ is distributed to predicate $q$ and predicate $\neg q$ which are "OR($\vee$)" operated with each other and $\neg p$ is "AND($\wedge $)" operated over all


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