Natural deduction for predicate logic -

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i've been stuck on particular predicate logic problem (using coq) long time. i've solved 30-40 predicate logic problems 1 can't figure out.

this problem: ~all x, (p(x) / (q(x) -> t(x))) -> ~all x, t(x).

or in box form

can send me in right direction? thanks!

edit:

this coq code problem:

variables p q t : d -> prop.  theorem pred_015 : ~all x, (p(x) \/ (q(x) -> t(x))) -> ~all x, t(x). proof. imp_i h.    qed.

it looks me using old version of coq. after adding missing declaration d, , replacing all forall, statement not provable. however, if had set of parentheses, goal provable. see following code:

variable d : set. variables p q t : d -> prop.  theorem pred_015 : (~forall x, (p(x) \/ (q(x) -> t(x)))) -> ~forall x, t(x). proof.

now, don't think should giving solution here, in public, it's quite easy if remember ~h defined h -> false.


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