Consider the following argument with premise $({\forall _x}P\left( x \right)) \vee Q\left( x \right))$ and conclusion $({\forall _x}P\left( x \right)) \wedge (\forall_xQ\left( x \right))$

(A) ∀_x (P(x) ∨ Q(x))

Premise

(B) P(c) ∨ Q(c)

Universal instantiation from (A)

(C) P(c)

Simplification from (B)

(D) ∀_x P(x)

Universal Generalization of (C)

(E) Q(c)

Simplification from (B)

(F) ∀_x Q(x)

Universal Generalization of (E)

(G) (∀_x P(x)) ∧ (∀_x Q(x))

Conjunction of (D) and (F)

Question

Question

This question was previously asked in

UGC NET Computer Science (Paper 2) 2020 Official Paper

Answer 1

The correct answer is option2

Explanation:

For C and E to be true, the assertion should be P(c) ∧ Q(c). Hence, this is not a correct inference

Additional Information

In predicate logic, universal generalization states that if $\vdash \!P(x)$ has been derived, then $\vdash \!\forall x\,P(x)$ can be derived.

(A) ∀_x (P(x) ∨ Q(x))	Premise
(B) P(c) ∨ Q(c)	Universal instantiation from (A)
(C) P(c)	Simplification from (B)
(D) ∀_x P(x)	Universal Generalization of (C)
(E) Q(c)	Simplification from (B)
(F) ∀_x Q(x)	Universal Generalization of (E)
(G) (∀_x P(x)) ∧ (∀_x Q(x))	Conjunction of (D) and (F)