The statement (p ⇒ q) ∨(p ⇒ r) is NOT equivalent to:

Question

The statement (p ⇒ q) ∨(p ⇒ r) is NOT equivalent to: (A) ( p ∧(∼ r )) ⇒ q (B) (∼ q) ⇒((∼ r) ∨ p) (C) p ⇒( q ∨ r ) (D) ( p ∧(∼ q )) ⇒ r

Tardigrade Admin · Accepted Answer

(p arrow q) ∨(p arrow r) (∼ p ∨ q) ∨(∼ p ∨ r) =∼ p ∨(q ∨ r) =p arrow(q ∨ r) ≡(3) text is true. Now (1) ( p ∧ ∼ r ) arrow q ∼( p ∧ ∼ r ) ∨ q =(∼ p ∨ r ) ∨ q =∼ p ∨( r ∨ q )= p arrow( q ∨ r ) (4) ( p ∧ ∼ q ) arrow r = p arrow( q ∨ r )

Q. The statement $(p \Rightarrow q) \lor (p \Rightarrow r)$ is NOT equivalent to:

$(p \land (\sim r)) \Rightarrow q$

$(\sim q) \Rightarrow ((\sim r) \lor p)$

$p \Rightarrow (q \lor r)$

$(p \land (\sim q)) \Rightarrow r$

Q. The statement (p⇒q)∨(p⇒r) is NOT equivalent to:

(p∧(∼r))⇒q

(∼q)⇒((∼r)∨p)

p⇒(q∨r)

(p∧(∼q))⇒r

(p→q)∨(p→r) (∼p∨q)∨(∼p∨r) =∼p∨(q∨r) =p→(q∨r)≡(3) is true. Now (1) (p∧∼r)→q ∼(p∧∼r)∨q=(∼p∨r)∨q =∼p∨(r∨q)=p→(q∨r) (4) (p∧∼q)→r=p→(q∨r)

Q. The statement $(p \Rightarrow q) \lor (p \Rightarrow r)$ is NOT equivalent to:

$(p \land (\sim r)) \Rightarrow q$

$(\sim q) \Rightarrow ((\sim r) \lor p)$

$p \Rightarrow (q \lor r)$

$(p \land (\sim q)) \Rightarrow r$