# [Agda] Re: with + continuation

Peter Divianszky divipp at gmail.com
Sat Nov 17 14:51:12 CET 2012

```On 17/11/2012 14:25, Andreas Abel wrote:
> On 17.11.12 1:07 PM, Peter Divianszky wrote:
>>> It would be nice if one could write the following in Agda:
>>>
>>> ---------------------
>>> module _ (f : (A → B) → C) (g : A → B) where
>>>
>>>    c : C
>>>    c withcont f
>>>    ... | a = g a
>>> ---------------------
>>>
>>> Given 'f : (A → B) → C', 'withcont f' would give a value of 'A' and it
>>> would replace the goal 'C' with 'B'.
>>>
>>> This could be useful in the following situation:
>>>
>>> ---------------------
>>> open import Data.List using (List; []; _∷_)
>>> open import Data.Empty using (⊥)
>>> open import Relation.Nullary using (Dec; yes; no)
>>> open import Relation.Binary using (Decidable)
>>> open import Relation.Binary.PropositionalEquality using (_≡_; refl)
>>>
>>> module Try {A : Set} {{eq : Decidable {A = A} _≡_}} where
>>>
>>>    infix 2 _∈_ _∈?_
>>>
>>>    data _∈_ (x : A) : List A → Set where
>>>      here : ∀ {xs} → x ∈ x ∷ xs
>>>      there : ∀ {y ys} → x ∈ ys → x ∈ y ∷ ys
>>>
>>>    _∈?_ : ∀ (x : A) ys → Dec (x ∈ ys)
>>>    x  ∈? []  = no λ ()
>>>    x  ∈? y ∷ ys with eq x y
>>>    .y ∈? y ∷ ys | yes refl = yes here
>>>    x  ∈? y ∷ ys | no ¬p with x ∈? ys
>>>    x  ∈? y ∷ ys | no ¬p | yes q = yes (there q)
>>>    x  ∈? y ∷ ys | no ¬p | no ¬q = no (f x refl)  where
>>>      f : ∀ a → x ≡ a → (a ∈ y ∷ ys) → ⊥
>>>      f .y e here = ¬p e
>>>      f a e (there r) with a | e
>>>      f a e (there r) | .x | refl = ¬q r
>>> ---------------------
>>>
>>> This works but I find the following definition of _∈?_ easier to read:
>>>
>> ---------------------
>>     _∈?_ : ∀ (x : A) ys → Dec (x ∈ ys)
>>     x  ∈? []  = no λ ()
>>     x  ∈? y ∷ ys with eq x y
>>     .y ∈? y ∷ ys | yes refl = yes here
>>     x  ∈? y ∷ ys | no ¬p with x ∈? ys
>>     x  ∈? y ∷ ys | no ¬p | yes q = yes (there q)
>>     x  ∈? y ∷ ys | no ¬p | no ¬q withcont no
>>     .y ∈? y ∷ ys | no ¬p | no ¬q | here    = ¬p refl
>>     x  ∈? y ∷ ys | no ¬p | no ¬q | there r = ¬q r
>> ---------------------
>
> Makes sense.  I'd call this `refine' or `with-apply' instead of `withcont'.

'without' would also make sense because 'no' is missing in the result :)

```