**Identify the error or errors in this argument that supposedly shows that if ∃xP (x) ∧ ∃xQ(x) is true then ∃x(P (x) ∧ Q(x)) is true. ∧**

The **Answer to the Question**

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**Here's the Solution to this Question**

The statement provided in the questions is

if$\exist x_p(n)\land \exist nQ(x)$ is true then

$\exist x(P(x)\land Q(n))$ is true

TO support this arguments, These arguments are given

a) ∃xP (x) ∨ ∃xQ(x) Premise

b) ∃xP (x) Simplification from (1)

c) P (c) Existential instantiation from (2)

d) ∃xQ(x) Simplification from (1)

e) Q(c) Existential instantiation from (4)

f) P (c) ∧ Q(c) Conjunction from (3) and (5)

g) ∃x(P (x) ∧ Q(x)) Existential generalization

In step(1), there is an error in the Premise, as dis-conjunction is used instead of conjunction.

In step(5), It cannot be assumed that the same c makes the p as well as Q true as it is mentioned that p is true for the same value of n and Q is also true for some value of x, but it is not provided that both of them are true for the same value of x.