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!
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Sign up to join this communityHow 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!
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 $$(\neg p \land q) \lor ( \neg p \land \neg q)$$
$$\Leftrightarrow (\neg p \lor \neg p) \land ( q \lor \neg q)\text{ (By Distributivite Law)}$$
$$\Leftrightarrow \neg p \land (q \lor \neg q) \text{ (By Idempotent Law)} $$
v
, @OGC?
$\endgroup$
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