Consider the following datatypes with bindings in Nominal Isabelle:
theory Example
imports "Nominal2.Nominal2"
begin
atom_decl vrs
nominal_datatype ty =
Tvar "vrs"
| Arrow x::vrs T::"ty" binds x in T
nominal_datatype trm =
Var "vrs"
| Abs x::"vrs" t::"trm" binds x in t
inductive
typing :: "trm ⇒ ty ⇒ bool" ("_ , _" [60,60] 60)
where
T_Abs[intro]: "(Abs x t) , (Arrow x T)"
equivariance typing
nominal_inductive typing done
lemma
assumes "(Abs x t), (Arrow y T)"
shows "x = y"
using assms
I want to prove that the two bindings appearing in the relation are equal. I see two ways an Isabelle user could help:
- If you know Nominal Isabelle is it possible to do this?
Unfortunately, it is not possible to prove the theorem that you are trying to prove. Here is a counterexample (the proofs were Sledgehammered):
theory Scratch
imports "Nominal2.Nominal2"
begin
atom_decl vrs
nominal_datatype ty =
Tvar "vrs"
| Arrow x::vrs T::"ty" binds x in T
nominal_datatype trm =
Var "vrs"
| Abs x::"vrs" t::"trm" binds x in t
inductive
typing :: "trm ⇒ ty ⇒ bool" ("_ , _" [60,60] 60)
where
T_Abs[intro]: "(Abs x t) , (Arrow x T)"
equivariance typing
nominal_inductive typing .
abbreviation s where "s ≡ Sort ''Scratch.vrs'' []"
abbreviation v where "v n ≡ Abs_vrs (Atom s n)"
lemma neq: "Abs (v 1) (Var (v 0)), Arrow (v (Suc (Suc 0))) (Tvar (v 0))"
(is "?a, ?b")
proof-
have a_def: "Abs (v 1) (Var (v 0)) = Abs (v (Suc (Suc 0))) (Var (v 0))"
(*Sledgehammered*)
by simp (smt Abs_vrs_inverse atom.inject flip_at_base_simps(3) fresh_PairD(2)
fresh_at_base(2) mem_Collect_eq nat.distinct(1) sort_of.simps trm.fresh(1))
from typing.simps[of ?a ?b, unfolded this, THEN iffD2] have
"Abs (v (Suc (Suc 0))) (Var (v 0)) , Arrow (v (Suc (Suc 0))) (Tvar (v 0))"
by auto
then show ?thesis unfolding a_def by clarsimp
qed
lemma "∃x y t T. x ≠ y ∧ (Abs x t), (Arrow y T)"
proof(intro exI conjI)
show "v 1 ≠ v (Suc (Suc 0))"
(*Sledgehammered*)
by (smt Abs_vrs_inverse One_nat_def atom.inject mem_Collect_eq n_not_Suc_n
sort_of.simps)
show "Abs (v 1) (Var (v 0)) , Arrow (v (Suc (Suc 0))) (Tvar (v 0))"
by (rule neq)
qed
end
- Otherwise, are the two occurrences of x in the rule T_Abs equal for the assistant or are they sort of bound variable with different identity?
I believe that you are thinking along the right lines and, hopefully, the example above will clarify any confusion that you might have. Generally, you could interpret the meaning of Abs x t1 = Abs y t2 as the alpha-equivalence of (λx. t1) and (λy. t2). Of course, (λx. t1) and (λy. t2) may be alpha equivalent without x and y being equal.