Answer to Question #190961 in Discrete Mathematics for Maaz

Question #190961

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.

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


1
Expert's answer
2021-05-11T13:22:14-0400

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 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.


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