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

y > 1/3x

y < = -x + 4

hope this : )

answer: translate the circles so they share a common center point and dilate circle y by a scale factor of 3.