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. 1. ∃xp (x) ∨ ∃xq (x) premise 2. ∃xp (x) simplification from (1) 3. p (c) existential instantiation from (2) 4. ∃xq (x) simplification from (1) 5. q (c) existential instantiation from (4) 6. p (c) ∧ q (c) conjunction from (3) and (5) 7. ∃x (p (x) ∧ q (x)) existential generalization

The mistake lies in steps 3 and 5.

In fact, you know that there exists some element that satisfies p (x), and some element that satisfies q (x), but you can't assume that the same element satisfies p (x) and q (x) at the same time.

So, you are only allowed to say

p (c) existential instantiation from (2)

q (d) existential instantiation from (4)

and so you can't do the conjunction in step 6 anymore.

Here's an example: if you follow this logic, you would say something like:

"There exists a person who is born in Italy and there exists a person who is born in the USA. Therefore, there exists a person who is born in both Italy and the USA"

which is clearly false: altought there exists a person for each proposition, they are not the same person.

This is even easier if q is the negation of p:

"There exists a number greater than 10, and there exists a number lesser than 10. Therefore, there exists a number that is greater than and lesser than 10 at the same time"

which is clearly absurd