$P$ and $Q$ are two propositions. Which of the following logical expressions are equivalent?
I. P∨ ~Q
II. ~ ( ~P∧Q)
III. (P∧Q)∨(P∧~ Q)∨( ~P∧ ~Q)
IV. (P∧Q)∨(P∧~ Q)∨( ~P∧Q)
I and II are present in all options so need to check.
For III and IV
$(P∧Q)∨(P∧\neg Q) \equiv P∧(QV\neg Q)$ (By distributive law)
$\quad \quad \equiv P∧T$
$\quad \quad \equiv P$
$P∨( \neg P∧\neg Q) \equiv P\vee \neg Q$ (By Absorption Law)
$P∨(¬P∧Q) \equiv P∨Q$ (By Absorption Law)
So Option B is correct.
I, II, III are logically equivalent.
I. P ∨ ~Q
II. ~ ( ~P ∧ Q) apply De Morgan's law. we get P ∨ ~Q
III. (P ∧ Q) ∨ (P ∧~ Q) ∨ ( ~P ∧ ~Q)
=[(P ∧ Q) ∨ (P ∧ ~Q)] ∨ ( ~P ∧ ~Q)
=[P(Q V ~Q)] ∨ ( ~P ∧ ~Q)
=P ∨ ( ~P ∧ ~Q)
= P ∨ ~Q
Hi @Anu , P + (~P~Q) , how you are deriving it to be P+~Q. Can you please explain ?
Thank you so much Arjun Sir!