How to define an inductive predicate inside locale? - isabelle

Question

Here is an example of a simple locale:
locale test =
fixes test_less_eq :: "'a ⇒ 'a ⇒ bool"
begin
inductive test_eq where
"test_less_eq x y ⟹ test_less_eq y x ⟹ test_eq x y"
end
It defines inductive test_eq. It can be defined using definition, but I need it to be an inductive predicate.
Then I define a trivial interpretation of the locale and try to use it:
interpretation interp: test "op <" .
inductive some_pred where
"interp.test_eq x y ⟹
some_pred x y"
code_pred [show_modes] some_pred .
The problem is that I get the following error for code_pred:
Type mismatch of predicate test.test_eq (trying to match ?'a
⇒ ?'a ⇒ bool and ('a ⇒ 'a ⇒ bool)
⇒ 'a ⇒ 'a ⇒ bool) in ?x1 < ?y1 ⟹
?y1 < ?x1 ⟹ interp.test_eq ?x1 ?y1
What is a cause of the error and how to fix it?

Answer 1

The predicate compiler has never been localized, i.e., it cannot directly deal with predicates that are defined inside a locale. There are two ways to make this work nevertheless.
Either, use global_interpretation with a defines clause to introduce a new constant for the predicate (plain interpretation only introduces an abbreviation). Then, you also have to re-declare the introduction rules to code_pred and prove the corresponding elimination rule.
global_interpretation interp: test "op <"
defines interp_test_eq = interp.test_eq .
declare interp.test_eq.intros[code_pred_intro]
code_pred interp_test_eq by(rule interp.test_eq.cases)
Or, leave the interpretation as is and re-declare the introduction rules of the internal constant to which the definition in the locale is mapped. This is <locale_name>.<predicate_name>, i.e., test.test_eq in your case. This works only if your locale has no assumption.
declare test.test_eq.intros[code_pred_intro]
code_pred test.test_eq by(rule test.test_eq.cases)