~ (exists x:D, ~ R x)->(forall y:D, R y)
I have worked on it for quite a long time, but it seems that I cannot use the left part of the implication well.
This is the first part of my code:
Parameter D: Set.
Parameter x: D.
Parameter y: D.
Parameter R: D->Prop.
Lemma b: ~(exists x:D, ~ (R x))->(forall y:D, (R y)).
Can anyone help me figure out how to write the rest of of the code?
Your question is a bit vague, as you don't specify what D and R are, and where you are stuck in your proof. Try providing a minimal working example, with an explicit fail tactic for where you're stuck in the proof.
In classical logic (the one you're use to in maths), as you have the excluded middle rule, you can always do a case analysis on whether something is true or false. In vanilla Coq, built for intuitionistic logic, it's not the case. Your result is actually not provable if the predicate R is not decidable (if it's not either true or false on every input : forall (x:D), R x \/ ~R x), if the type D is not empty.
Try adding the decidability of R as an hypothesis and reprove it. It should follow more or less this structure (the key being the case analysis on whether (R y) is true or false) :
Parameter D: Set.
Parameter R: D -> Prop.
Lemma yourGoal :
(forall x, R x \/ ~ R x) -> (* Decidability of R *)
~ ( exists x, ~ (R x) )->
forall y, (R y).
Proof.
intros Hdec Hex y. (* naming the hypothesis for convenience *)
specialize (Hdec y).
destruct Hdec as [H_Ry_is_true | H_Ry_is_false]. (* case analysis, creates two goals *)
+ (* (R y) is true, which is our goal. *)
assumption.
+ (* (R y) is false, which contradicts Hex *)
exfalso. (* transform your goal into False *)
apply Hex.
(* should be easy from here, using the [exists] tactic *)
Qed.
Ps: this exact result (and its link with excluded middle) is mentioned in Software foundations, which is a great resource to learn Coq and logic : https://softwarefoundations.cis.upenn.edu/lf-current/Logic.html#not_exists_dist