Isabelle: Evaluating formula with Quantifiers - isabelle

Question

Why does the following work:
fun f :: "nat ⇒ bool" where
"f _ = (True ∨ (∀x. x))"
But this fails
fun g :: "nat ⇒ bool" where
"g _ = (True ∨ (∀a. True))"
with
Additional type variable(s) in specification of "g_graph": 'a
Specification depends on extra type variables: "'a"
The error(s) above occurred in "test.g_sumC_def"
The error(s) above occurred in definition "g_sumC_def":
"g_sumC ≡ λx. THE_default undefined (g_graph TYPE('a) x)"
Similarly, the following succeeds,
value "True ∨ (∀x. x)"
but this fails
value "True ∨ (∀x. True)"
with
Wellsortedness error:
Type 'a not of sort enum
Cannot derive subsort relation {} < enum

Answer 1

The short answer is: In your first function definition type inference easily infers that x is of type bool, while in your second definition, the bound variable a is not used anywhere else and thus its type is arbitrary ('a). Which is what Additional type variable(s) in specification ... expresses.
If you constrain the type of a explicitly, e.g.,
fun g :: "nat ⇒ bool" where
"g _ = (True ∨ (∀a::bool. True))"
the function definition is accepted.
A longer answer: Since the definition of g is not recursive you could turn it into using definition instead of fun. Then your first attempt does not fail completely but the result might surprise you. After
definition g :: "nat ⇒ bool" where
"g _ = (True ∨ (∀a. True))"
the type of g is 'a itself => nat => bool instead of the intended nat => bool. The reason is the same as for the failure of fun before. Since a is of arbitrary type, this additional type has to be recorded in the type of g, which is done by introducing an additional dummy argument which just states this additional type explicitly. Here 'a itself is a type whose constructor TYPE(...) -- taking a type as argument -- allows us to encode type information on the term level. E.g.,
TYPE('a) :: 'a itself
TYPE(bool) :: bool itself
TYPE(nat) :: nat itself
Then g TYPE(nat) is the version of g where a is fixed to be of type nat.
Concerning your value statements, the reason for the second one to fail is not really related to the above. In the first statement the universal quantifier binds a variable of type bool whose values can be enumerated explicitly and thus a result can be computed by considering all those values. In contrast, in your second statement the bound variable x is of an arbitrary type 'a whose values cannot be enumerated explicitly.

Answer 2

The following fails:
fun f where "f _ = (∀a. True)"
because the type of a has hidden polymorphism (i.e., there is a type variable inside your function's body that is not present in the function's type signature), which upsets the function package's internal proofs.
If you explicitly give a a type as so:
fun f where "f _ = (∀a::bool. True)"
or is you give a a type that is also in the function's type signature, as so:
fun f where "f _ = (∀a::bool. True)"
the function definition succeeds. Your example:
fun f where "f _ = (∀x. x)"
succeeds, because x is forced to be type bool.
As for your value commands, Isabelle attempts to generate executable code for your expression, and hence needs to not only know the type of your for-all statements, but also be able to enumerate all possible values of it, so that it can test them all. Types such as bool work fine, but type variables like 'a or infinite types such as nat are not enumerable, and hence Isabelle cannot generate code for them.