Answer question 3.1

amethyst
Shad Amethyst 11 months ago
parent 518b4cb20c
commit ef2a8c2b75

@ -2,6 +2,19 @@ From stdpp Require Import gmap base relations.
From iris Require Import prelude.
From semantics.ts.systemf Require Import lang notation types tactics.
(*
Exercise 3.1:
With V[A->B] = {v | v', v' V[A] -> v v' E[B]}
The set V[A->B] is equal to:
V[A->B] = {v | v', v' V[A], w, v v' | w, w V[B]}
According to big step semantics, `v v' | w` inverts to `v | λx.e` and `e[v'/x] | w`.
Assuming that `v` and `v'` are closed, we have `e[v'/x]` closed, and thus `e[v'/x] E[B]`.
Since `v` is a value and `v | λx.e`, we have `v = λx.e`.
We thus have `{v | v', v' V[A] -> v v' E[B]} = {λx.e | v', v' V[A] -> e[v'/x] E[B]}`,
after which we can carry out our proof as before.
*)
(** Exercise 3 (LN Exercise 22): Universal Fun *)
Definition fun_comp : val :=
@ -96,5 +109,3 @@ Proof.
(* Should be solved by reflexivity. *)
(* TODO: exercise *)
Admitted.

Loading…
Cancel
Save