考虑以下在Nominal Isabelle中具有绑定的数据类型:
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
我想证明关系中出现的两个绑定是相等的。我认为Isabelle用户可以通过两种方式提供帮助:
- 如果你知道Nominal Isabelle,有可能做到这一点吗
- 否则,x在规则T_Abs中的两次出现对于辅助是相等的,还是它们是具有不同恒等式的有界变量
- 如果你知道Nominal Isabelle,有可能做到这一点吗
不幸的是,不可能证明你试图证明的定理。这里有一个反例(证明是Sledgehammer
ed(:
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
- 否则,规则T_Abs中出现的两个x是否等于辅助变量还是它们是不同的绑定变量身份
我相信您的思路是正确的,希望上面的例子能澄清您可能存在的任何困惑。一般来说,你可以将Abs x t1 = Abs y t2
的含义解释为(λx. t1)
和(λy. t2)
的α等价。当然,在x
和y
不相等的情况下,(λx. t1)
和(λy. t2)
可以是α等价的。