How to prove this natural deduction?

Question

I'm trying to prove this formula but its really hard.. Here is the formula:

¬∃x.(P(x)∧R(x)) Premisse
¬∃x.(S(x)∧¬R(x)) Premisse
∀x.(A(x)→P(x)) Premisse

∀x.(A(x)→S(x)) Conclusion

I'm actual in this step:

Any one knows how to proceed?

Answer 1

• Assume A(x) is true.
• By Premise 3 P(x) must be true.
• By Premise 1 R(x) must be false (otherwise, P(x) and R(x) would be both true)
• By Premise 2 S(x) must be false (otherwise, S(x) and ¬R(x) would be both true)
• Therefore ¬S(x) is true

So, A(x) → ¬S(x) and given that x was arbitrarily chosen:

∀x.(A(x) → ¬S(x))

and the conclusion you were given is wrong.