How can I prove the following free theorem with the plugin Paramcoq?
Lemma id_free (f : forall A : Type, A -> A) (X : Type) (x : X), f X x = x.
If it is not possible, then what is the purpose of this plugin?
Paramcoq: Free theorems in Coq
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The plugin can generate the statement of parametricity for any type. You will still need to then declare it as an axiom or an assumption to actually use it: