Isabelle proving with translation issue - isabelle

Question

I had defined a few translations like this:
consts
"time" :: "i"
"sig" :: "i ⇒ i"
"BaseChTy" :: "i"
syntax
"time" :: "i"
"sig" :: "i ⇒ i"
translations
"time" ⇌ "CONST int"
"sig(A)" ⇌ "CONST int → A"
Then, I want to prove a theorem like this:
theorem sig_mono: "⟦ A ⊆ B ⟧ ⟹ sig(A) ⊆ sig(B)"
It should be a very simple theorem, and should be proved with theorem Pi_mono in a single step:
thm Pi_mono
?B ⊆ ?C ⟹ ?A → ?B ⊆ ?A → ?C
So I did it like this:
theorem sig_mono: "⟦ A ⊆ B ⟧ ⟹ sig(A) ⊆ sig(B)"
apply(drule Pi_mono[of _ _ "time"])
(*Output:
goal (1 subgoal):
1. sig(A) ⊆ sig(B) ⟹ sig(A) ⊆ sig(B)
*)
apply(simp)
(*Output:
Failed ...
*)
Since the premise has become the same as the goal, it should be proved immediately, but it didn't. May I know have I done anything wrong in the translation definition?
I tried to change the theorem to:
theorem sig_mono: "⟦ A ⊆ B ⟧ ⟹ (time → A) ⊆ (time → B)"
(*Output:
goal (1 subgoal):
1. A ⊆ B ⟹ sig(A) ⊆ sig(B)
*)
apply(drule Pi_mono[of _ _ "time"])
(*Output:
goal (1 subgoal):
1. sig(A) ⊆ sig(B) ⟹ sig(A) ⊆ sig(B)
*)
apply(simp)
(*Output:
Success ...
*)
Then it works immediately, but shouldn't the translation will make them to be the same thing?
Update:
Thanks for Mathias Fleury reply, I tried to do a simplify trace, and it shows something like this:
theorem sig_mono: "⟦ A ⊆ B ⟧ ⟹ sig(A) ⊆ sig(B)"
using [[show_sorts]] apply(drule Pi_mono[of _ _ "time"])
using [[simp_trace]] apply(simp)
oops
(*
Output:
[1]SIMPLIFIER INVOKED ON THE FOLLOWING TERM:
sig(A::i) ⊆ sig(B::i) ⟹ sig(A) ⊆ sig(B)
[1]Adding rewrite rule "??.unknown":
sig(A::i) ⊆ sig(B::i) ≡ True
*)
while the time -> A version shows:
theorem sig_mono: "⟦ A ⊆ B ⟧ ⟹ time → A ⊆ time → B"
using [[show_sorts]] apply(drule Pi_mono[of _ _ "time"])
using [[simp_trace]] apply(simp)
oops
(*
Output:
[1]SIMPLIFIER INVOKED ON THE FOLLOWING TERM:
sig(A::i) ⊆ sig(B::i) ⟹ sig(A) ⊆ sig(B)
[1]Adding rewrite rule "??.unknown":
sig(A::i) ⊆ sig(B::i) ≡ True
[1]Applying instance of rewrite rule "??.unknown":
sig(A::i) ⊆ sig(B::i) ≡ True
[1]Rewriting:
sig(A::i) ⊆ sig(B::i) ≡ True
*)
Why can this time version can apply the instance of rewrite rule to continue to the proof, but the original one does not?

Answer 1

Thanks to the imports you mentioned in you comment (thanks), I could reproduce the problem. The issue is the translation, you need to do something like
syntax
"sig" :: "i ⇒ i" (‹sig(_)›)
translations
"sig(A)" == "CONST int → A"
theorem sig_mono: "⟦ A ⊆ B ⟧ ⟹ sig(A) ⊆ sig(B)"
apply(rule Pi_mono)
apply assumption
done
Just to expand on my comment and explain how I found that the problem is the translation. I looked at the unification failure:
theorem ⟦ A ⊆ B ⟧ ⟹ time → A ⊆ time → B
supply[[unify_trace_failure]]
apply (rule PI_mono)
The error message tells that sig and Pi are not unifiable. This is already strange. To be certain that the problem comes from the translation, I looked at the underlying term:
ML ‹#{print}#{term ‹sig(A)›}›
It shows the underlying term and we can see that the translation is not working and I looked at other translations in the library to fix issue.