@hammersklavier,
1. Ǝx(Gx -> Px)
2. ∀x(Px -> Ex)
ergo (Ǝx)(Gx -> Px)
First problem is that this argument begs the question. Assumption 1 is your conclusion, unless the parenthesis around your Ǝx makes the conclusion different from 1. But from my understanding, it shouldn't. All you really need to do is use a reiteration rule or a copy rule.
Do you mean to say?
1. Ǝx(Gx -> Px)
2. ∀x(Px -> Ex)
ergo (Ǝx)(Gx -> Ex)
---
1. Ǝx(Gx -> Px) | A
2. ∀x(Px -> Ex) | A
3. Negated conclusion | Provisional Assumption
4. Ax-(Gx -> Ex) | Quantifier Exchange, 3
5. Ga -> Pa | Existential Out, 1
6. -(Ga -> Ea) | Universal Out, 4
7. Pa -> Ea | Universal Out, 2
8. Ga . -Ea | Arrow Rule, 6
9. Pa | Modus Ponens, 5, 8 (one of the conjunctions)
10. Ea | Modus Ponens, 7, 9
11. -Ea | Conjunction out, 8
12. Ea . -Ea | Conjunction in, 10, 11
ergo (Ǝx)(Gx -> Ex) | Reductio ad absurdum or indirect proof
But even then, I thought existential statements were supposed to have conjunction symbols between the terms, not implication symbols.