How to define a subformula of an inductively defined type in Agda?

强颜欢笑 提交于 2021-01-29 08:11:22

问题


I'm trying to define a simple predicate to determine if a formula is a subformula of a given formal over a simple inductively defined syntax. I'm running into a few, presumably simple, problems.

(i) I'd rather use a parameterized module with a given type A. How can one import the information that A is a set, in the sense that A has decideable equality? If this can't be done, what are some workarounds? This is why I have Nat instead.

(ii) Is the t1 ≡ (t2 // t3) (and similairly defined) predicates isSubFormula the proper way to deal with equal subformulas? If not, how else does one easily do this? I've also considered writing a predicate for equalFmla and then making a global subformula predicate with isSubFormula ⊎ equalFmla but am not sure if this is so clean.

(iii) Why do the last three lines highlight blue when I pattern match internal to the first one? How can I fix this

(iv) Why won't the {!Data.Nat._≟_ n1 n2 !} refine below?

module categorial1 (A : Set) where

open import Data.Nat using (ℕ)
open import Data.Empty
open import Data.Sum
open import Relation.Binary.PropositionalEquality

-- type symbols
data tSymb : Set where
  -- base : A → tSymb
  base : ℕ → tSymb
  ~ : tSymb → tSymb
  _\\_ : tSymb → tSymb → tSymb
  _//_ : tSymb → tSymb → tSymb

-- A needs a decideable notion of equality
isSubFormula : tSymb → tSymb → Set
-- isSubFormula y (base x) = {!!}
isSubFormula (base n1) (base n2) = {!Data.Nat._≟_ n1 n2 !}
isSubFormula (~ t1) (base x) = ⊥
isSubFormula (t1 \\ t2) (base x) = ⊥
isSubFormula (t1 // t2) (base x) = ⊥
isSubFormula t1 (~ t2) = t1 ≡ (~ t2) ⊎ isSubFormula t1 t2
isSubFormula t1 (t2 \\ t3) = t1 ≡ (t2 \\ t3) ⊎ isSubFormula t1 t2 ⊎ isSubFormula t1 t3
isSubFormula t1 (t2 // t3) = t1 ≡ (t2 // t3) ⊎ isSubFormula t1 t2 ⊎ isSubFormula t1 t3


回答1:


That's quite a lot of different questions; Stack Overflow works better if you post each question separately.

Nevertheless, here are some answers:

(i) I'd rather use a parameterized module with a given type A. How can one import the information that A is a set, in the sense that A has decideable equality?

You can parametrize your module by the decision procedure, i.e. something like this:

open import Relation.Binary
open import Relation.Binary.PropositionalEquality

module categorical1 {A : Set} (_≟_ : Decidable (_≡_ {A = A})) where

Note that we import Relation.Binary (for Decidable) and R.B.PropositionalEquality (for _≡_) before the module header, since the parameter types of our module depend on these definitions.

(iii) Why do the last three lines highlight blue when I pattern match internal to the first one? How can I fix this

This is Agda warning you that those clauses won't hold definitionally, because they depend on the previous clauses not matching on t1.

(iv) Why won't the {!Data.Nat. n1 n2 !} refine below?

Because isSubFormula returns a Set, but n1 ≟ n2 returns a Dec _≡_.

I think there is confusion in your code about whether isSubFormula is supposed to be a proposition or a decision procedure. If it returns a Set, it means it is a proposition. You can write that without decidable equality since there is nothing to decide -- isSubFormula-ness of two base values means they are equal:

isSubFormula : tSymb → tSymb → Set
isSubFormula (base n1) (base n2) = n1 ≡ n2

If you want a decision procedure, you could do it "boolean-blindly" by instead writing isSubFormula : tSymb → tSymb → Bool, or by keeping isSubFormula as a proposition and writing decSubFormula : Decidable isSubFormula.

Note also that you only need decidable equality on A if you are trying to decide isSubFormula, so you can do

module categorical1 (A : Set) where

... 

isSubFormula : tSymb → tSymb → Set

decSubFormula : Decidable (_≡_ {A = A}) → Decidable isSubFormula
decSubFormula _≟A_ = ? -- You can use _≟A_ here to decide equality of A values


来源:https://stackoverflow.com/questions/64362886/how-to-define-a-subformula-of-an-inductively-defined-type-in-agda

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