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{-# OPTIONS --cubical-compatible #-}
module Padova2025.ProvingBasics.Equality.General where

General results on equality

open import Padova2025.ProvingBasics.Equality.Base

Exercise: Symmetry

In blackboard mathematics, we have the basic result that x = y implies y = x. In Agda, we have a function which transforms values of type x ≡ y into values of type y ≡ x. Fill in this hole:

sym : {A : Set} {x y : A} → x ≡ y → y ≡ x
sym = {!!}
  1. First introduce a variable to the left of the = symbol, i.e. have the line read sym p = ?. Then reload the file using C-c C-l, else the hole will not know about the variable p.
  2. Press C-c , to ask Agda to print a summary of the situation.
  3. Then do a case split on p.
  4. Agda then recognizes that the only possibility for p is refl. Press C-c , again to request a new printout of the situation. Notice that the situation has simplified.
  5. Fill in the hole with refl, press C-c C-SPACE and then reload the file.

Exercise: Congruence

In blackboard mathematics, we have the basic result that an equation remains true if the same operation is applied to its two sides: If x = y, then also f(x) = f(y). In Agda, we have a function which transformed values of type x ≡ y into values of type f x ≡ f y:

cong : {A B : Set} {x y : A} → (f : A → B) → x ≡ y → f x ≡ f y
cong = {!!}

Exercise: Transitivity

Fill in this hole, thereby proving that equality is transitive.

trans : {A : Set} {x y z : A} → x ≡ y → y ≡ z → x ≡ z
trans = {!!}

Exercise: Pointwise equality

Prove that equal functions have equal values. (The converse is a principle known as “function extensionality” which can be postulated in Agda but is not assumed by default.)

equal→pwequal : {A B : Set} {f g : A → B} {x : A} → f ≡ g → f x ≡ g x
equal→pwequal = {!!}

Exercise: Identity of indiscernibles

Identical values have all their properties in common: If F : A → Set is a property of elements of a type A (for instance, F might be the predicate Even from before) and if x and y are identical elements, then F x should imply F y.

subst : {A : Set} {x y : A} → (F : A → Set) → x ≡ y → F x → F y
subst = {!!}

Exercise: Congruence for functions with two parameters

cong₂
  : {A B C : Set} {x x' : A} {y y' : B}
  → (f : A → B → C) → x ≡ x' → y ≡ y' → f x y ≡ f x' y'
cong₂ = {!!}

Inequality

We can introduce inequality as follows.

open import Padova2025.ProvingBasics.Negation

infix 4 _≢_
_≢_ : {X : Set} → X → X → Set
a ≢ b = ¬ (a ≡ b)

Inequality is symmetric:

≢-sym : {X : Set} {a b : X} → a ≢ b → b ≢ a
≢-sym = {!!}

Pointwise equality

We will also use the notion that two functions have the same values, called pointwise equality. (We will later discuss the relation of this notion to the more basic notion that two functions are identical.)

infix 4 _≗_
_≗_ : {A B : Set} → (A → B) → (A → B) → Set
f ≗ g = (x : _) → f x ≡ g x

TODO: with in exercise

With abstraction

In some cases one needs to pattern match over a variable which depends on the inputs of a function. To do this, one can either introduce an auxiliary function with where or let and do another pattern matching there or one can use the with syntactic sugar. The latter is subject of the following section. 
{-# BUILTIN EQUALITY _≡_ #-}
module WithAbstraction where
  open import Padova2025.ProgrammingBasics.Booleans
  open import Padova2025.ProgrammingBasics.Lists
  open import Padova2025.ProgrammingBasics.Naturals.Base
A very simple example is the filter function, which is also shown in the Agda documentation. The filter function takes a function from an abstract type to Bool (a predicate) and a list and drops all elements of the list for which the predicate is false. Let’s first do that by avoiding syntactic sugar: 
  filter' : {A : Set} → (A → Bool) → List A → List A
  filter' f [] = []
  filter' {A} f (x ∷ xs) = aux (f x)
    where
      aux : Bool → List A
      aux false = filter' f xs
      aux true  = x ∷ filter' f xs

The induction case needs the function aux to pattern match over f x. This is very much boiler plate code, i.e. the function type and the function definition with its case splitting, which’s only purpose is to pattern match f x. We also have to make the implicit argument A : Set visible to be able to write down the type signature of aux.

Now we will introduce the with keyword to avoid this stuff. 
  filter : {A : Set} → (A → Bool) → List A → List A
  filter f [] = []
  filter f (x ∷ xs) with f x
  ... | false = filter f xs
  ... | true  = x ∷ filter f xs

This filter implementation is very compact. The with removes the necessity of introducing an auxiliary function. The ... in front are short for filter f (x ∷ xs), i.e. we also avoid repetition here. Observe that there is no = anymore between filter f (x ∷ xs) and with. This = is moved behind the false and the true in the case splitting below and tells us what filter f (x ∷ xs) should be in either case.

The following function demonstrates the with abstraction in a slightly trivial case: 
  eq-to-bool : {A : Set} → (x y : A) → x ≡ y → Bool
  eq-to-bool x y refl = true
The same could be done via with an pattern matching over p. The difference is, that after the with also a function of p would be possible. So it is more general than the original definition: 
  eq-to-bool' : {A : Set} → (x y : A) → x ≡ y → Bool
  eq-to-bool' x y p with p
  ... | refl = true
The aux function has the same purpose as the with and may also called with a function of p: 
  eq-to-bool'' : {A : Set} → (x y : A) → x ≡ y → Bool
  eq-to-bool'' x y p = aux p
    where
      aux : x ≡ y → Bool
      aux refl = true
Also nested withs are possible 
  show-with-nested : {A : Set} → (x y : List A) → Bool
  show-with-nested x y with length x | length y
  ... | succ (succ zero)  | succ zero  = true
  ... | _                 | _          = false
Notice, that the order of the with abstractions matter and also notice that an overall nested with is different to an additional with in one of the cases. with is particularily useful for constructing counter examples in negations. The following example is slightly artificial: 
  ex : (y : Bool) → y ≡ false → ((x : Bool) → x ≡ true) → ⊥
  ex y refl f with f y
  ... | ()
A with abstraction in a function definition makes a with abstraction necessary in a proof about this function. The following example about the idempotence of filter is also shown in the documentation and it is slightly more complicated than anticipated: 
  filter-idem : {A : Set} →
                (p : A → Bool) → (xs : List A) →
                (filter p (filter p xs)) ≡ (filter p xs)
  filter-idem f [] = refl

The base case is trivial.

The necessary type in the induction case is filter f (filter f (x ∷ xs) | f x) ≡ (filter f (x ∷ xs) | f x) which means that we need a with abstraction over f x. Notice, that on the left-hand side only the inner filter has an additional | f x part. This means Agda is stuck here and does not know whether the inner filtered list is filter f xs or x ∷ (filter f xs), because only in the x ∷ (filter f xs) case the with abstraction is necessary. 
  filter-idem f (x ∷ xs) with f x in eq
For the filter-idem theorem it is also necessary to introduce another mechanism in the with abstraction, namely in. This will be explained below. 
  ... | false = filter-idem f xs

The false case is trivial, because as per definition of filter, filter is called only with the rest of the list without the first element. This leads to collapse of both sides of the equation to the form of filter-idem f xs.

The true case is not so trivial. The documentation shows that the induction step could be proven by using a rewrite, 
--  ... | true rewrite eq = cong (x ∷_) (filter-idem f xs)

but the rewrite needs to be explained: With in after the f x we generate an equation eq in every with clause, in our case eq : f x ≡ false and eq : f x ≡ true. For this we needed the compiler pragma {-# BUILTIN EQUALITY ≡ #-} above. In the false case eq is not needed, because we can trivially use the induction hypothesis as the element x is removed. The true case is more complicated: The inner expression in (filter f (x ∷ xs) | f x) is (filter f (x ∷ xs) | true) and collapses to x ∷ filter f xs. But to complete the proof, we need x ∷_ outside of it, but we only have (filter f (x ∷ filter f xs) | f x).

We are in the true case of the with abstraction but have to re-establish the f x is still true. Therefore, we have to perform another with. This is due to the fact that the outer f x is only seen after we have established that the inner f x is true and have the x ∷ xs case in filter. f x cannot be generalized over (which means substituted by a variable where we may case split), since it already has the value true. Therefore, we need to use a dot pattern .true. But from this alone, Agda is not convinced that this is actually true as dot patterns are only for documentation purposes. (If we would have written bla instead of .true, the goal view would have be shown bla = true as a constraint.) We have to further with-abstract over eq with the case refl simultaneously. 
  ... | true with f x  | eq
  ... |          .true | refl = cong (x ∷_) (filter-idem f xs)

This construction appears quite often in proofs and therefore is abbreviated by rewrite above.

Let us switch the with ... in ... syntactic sugar on.