16. Functions in Lean

16.1. Functions and Symbolic Logic

Let us now consider functions in formal terms. Even though we have avoided the use of quantifiers and logical symbols in the definitions in the last chapter, by now you should be seeing them lurking beneath the surface. That fact that two functions \(f, g : X \to Y\) are equal if and only if they take the same values at every input can be expressed as follows:

\[\forall x \in X \; (f(x) = g(x)) \leftrightarrow f = g\]

This principle is a known as function extensionality, analogous to the principle of extensionality for sets, discussed in Section 12.1. Recall that the notation \(\forall x \in X \; P(x)\) abbreviates \(\forall x \; (x \in X \to P(x))\), and \(\exists x \in X \; P(x)\) abbreviates \(\exists x \; (x \in X \wedge P(x))\), thereby relativizing the quantifiers to \(A\).

We can avoid set-theoretic notation if we assume we are working in a logical formalism with basic types for \(X\) and \(Y\), so that we can specify that \(x\) ranges over \(X\). In that case, we will write instead

\[\forall x : X \; (f(x) = g(x) \leftrightarrow f = g)\]

to indicate that the quantification is over \(X\). Henceforth, we will assume that all our variables range over some type, though we will sometimes omit the types in the quantifiers when they can be inferred from context.

The function \(f\) is injective if it satisfies

\[\forall x_1, x_2 : X \; (f(x_1) = f(x_2) \to x_1 = x_2),\]

and \(f\) is surjective if

\[\forall y : Y \; \exists x : X \; f(x) = y.\]

If \(f : X \to Y\) and \(g: Y \to X\), \(g\) is a left inverse to \(f\) if

\[\forall x : X \; g(f(x)) = a.\]

Notice that this is a universal statement, and it is equivalent to the statement that \(f\) is a right inverse to \(g\).

Remember that in logic it is common to use lambda notation to define functions. We can denote the identity function by \(\lambda x \; x\), or perhaps \(\lambda x : X \; x\) to emphasize that the domain of the function is \(X\). If \(f : X \to Y\) and \(g : Y \to Z\), we can define the composition \(g \circ f\) by \(g \circ f = \lambda x : X \; g(f(x))\).

Also remember that if \(P(x)\) is any predicate, then in first order logic we can assert that there exists a unique \(x\) satisfying \(P(x)\), written \(\exists! x \; P(x)\), with the conjunction of the following two statements:

  • \(\exists x \; P(x)\)
  • \(\forall x_1, x_2 \; (P(x_1) \wedge P(x_2) \to x_1 = y_1)\)

Equivalently, we can write

\[\exists (P(x) \wedge \forall x' \; (P(x') \to x' = x)).\]

Assuming \(\exists! x \; P(x)\), the following two statements are equivalent:

  • \(\exists x \; (P(x) \wedge Q(x))\)
  • \(\forall x \; (P(x) \to Q(x))\)

and both can be taken to assert that “the \(x\) satisfying \(P\) also satisfies \(Q\).”

A binary relation \(R\) on \(X\) and \(Y\) is functional if it satisfies

\[\forall x \; \exists! y \; R(x,y).\]

In that case, a logician might use iota notation,

\[f(x) = \iota y \; R(x, y)\]

to define \(f(x)\) to be equal to the unique \(y\) satisfying \(R(x,y)\). If \(R\) satisfies the weaker property

\[\forall x \; \exists y \; R(x,y),\]

a logician might use the Hilbert epsilon to define a function

\[f(x) = \varepsilon y \; R(x, y)\]

to “choose” a value of \(y\) satisfying \(R(x, y)\). As we have noted above, this is an implicit use of the axiom of choice.

16.2. Second- and Higher-Order Logic

In contrast to first-order logic, where we start with a fixed stock of function and relation symbols, the topics we have been considering in the last few chapters encourage us to consider a more expressive language with variables ranging over functions and relations as well. For example, saying that a function \(f : X \to Y\) has a left-inverse implicitly involves a quantifying over functions,

\[\exists g \; \forall x \; g(f(x)) = x.\]

The theorem that asserts that if any function \(f\) from \(X\) to \(Y\) is injective then it has a left-inverse can be expressed as follows:

\[\forall x_1, x_2 \; (f(x_1) = f(x_2) \to x_1 = x_2) \to \exists g \; \forall x \; g(f(x)) = x.\]

Similarly, saying that two sets \(X\) and \(Y\) have a one-to-one correspondence asserts the existence of a function \(f : X \to Y\) as well as an inverse to \(f\). For another example, in Section 15.4 we asserted that every functional relation gives rise to a corresponding function, and vice-versa.

What makes these statements interesting is that they involve quantification, both existential and universal, over functions and relations. This takes us outside the realm of first-order logic. One option is to develop a theory in the language of first-order logic in which the universe contains functions, and relations as objects; we will see later that this is what axiomatic set theory does. An alternative is to extend first-order logic to involve new kinds of quantifiers and variables, to range over functions and relations. This is what higher-order logic does.

There are various ways to go about this. In view of the relationship between functions and relations described above, one can take relations as basic, and define functions in terms of them, or vice-versa. The following formulation of higher-order logic, due to the logician Alonzo Church, follows the latter approach. It is sometimes known as simple type theory.

Start with some basic types, \(X, Y, Z, \ldots\) and a special type, \(\mathrm{Prop}\), of propositions. Add the following two rules to build new types:

  • If \(U\) and \(V\) are types, so is \(U \times V\).
  • If \(U\) and \(V\) are types, so is \(U \to V\).

The first intended to denote the type of ordered pairs \((u, v)\), where \(u\) is in \(U\) and \(v\) is in \(V\). The second is intended to denote the type of functions from \(U\) to \(V\). Simple type theory now adds the following means of forming expressions:

  • If \(u\) is of type \(U\) and \(v\) is of type \(V\), \((u, v)\) is of type \(v\).
  • If \(p\) is of type \(U \times V\), then \((p)_1\) is of type \(U\) and \((p)_2\) if of type \(V\). (These are intended to denote the first and second element of the pair \(p\).)
  • If \(x\) is a variable of type \(U\), and \(v\) is any expression of type \(V\), then \(\lambda x \; v\) is of type \(U \to V\).
  • If \(f\) is of type \(U \to V\) and \(u\) is of type \(U\), \(f(u)\) is of type \(V\).

In addition, simple type theory provides all the means we have in first-order logic — boolean connectives, quantifiers, and equality — to build propositions.

A function \(f(x, y)\) which takes elements of \(X\) and \(Y\) to a type \(Z\) is viewed as an object of type \(X \times Y \to Z\). Similarly, a binary relation \(R(x,y)\) on \(X\) and \(Y\) is viewed as an object of type \(X \times Y \to \mathrm{Prop}\). What makes higher-order logic “higher order” is that we can iterate the function type operation indefinitely. For example, if \(\mathbb{N}\) is the type of natural numbers, \(\mathbb{N} \to \mathbb{N}\) denotes the type of functions from the natural numbers to the natural numbers, and \((\mathbb{N} \to \mathbb{N}) \to \mathbb{N}\) denotes the type of functions \(F(f)\) which take a function as argument, and returns a natural number.

We have not specified the syntax and rules of higher-order logic very carefully. This is done in a number of more advanced logic textbooks. The fragment of higher-order logic which allows only functions and relations on the basic types (without iterating these constructions) is known as second-order logic.

These notions should seem familiar; we have been using these constructions, with similar notation, in Lean. Indeed, Lean’s logic is an even more elaborate and expressive system of logic, which fully subsumes all the notions of higher-order logic we have discussed here.

16.3. Functions in Lean

The fact that the notions we have been discussing have such a straightforward logical form means that it is easy to define them in Lean. The main difference between the formal representation in Lean and the informal representation above is that, in Lean, we distinguish between a type X and a subset A : set X of that type.

In Lean’s library, composition and identity are defined as follows:

variables {X Y Z : Type}

def comp (f : Y  Z) (g : X  Y) : X  Z :=
λx, f (g x)

infixr  `  ` := comp

def id (x : X) : X :=
x

Ordinarily, we use funext (for “function extensionality”) to prove that two functions are equal.

example (f g : X  Y) (h :  x, f x = g x) : f = g :=
funext h

But Lean can prove some basic identities by simply unfolding definitions and simplifying expressions, using reflexivity.

lemma left_id (f : X  Y) : id  f = f := rfl

lemma right_id (f : X  Y) : f  id = f := rfl

theorem comp.assoc (f : Z  W) (g : Y  Z) (h : X  Y) :
  (f  g)  h = f  (g  h) := rfl

theorem comp.left_id (f : X  Y) : id  f = f := rfl

theorem comp.right_id (f : X  Y) : f  id = f := rfl

We can define what it means for \(f\) to be injective, surjective, or bijective:

def injective (f : X  Y) : Prop :=  x₁ x₂⦄, f x₁ = f x₂  x₁ = x₂

def surjective (f : X  Y) : Prop :=  y,  x, f x = y

def bijective (f : X  Y) := injective f  surjective f

Marking the variables x₁ and x₂ implicit in the definition of injective means that we do not have to write them as often. Specifically, given h : injective f, and h₁ : f x₁ = f x₂, we write h h₁ rather than h x₁ x₂ h₁ to show x₁ = x₂.

We can then prove that the identity function is bijective:

theorem injective_id : injective (@id X) :=
assume x₁ x₂,
assume H : id x₁ = id x₂,
show x₁ = x₂, from H

theorem surjective_id : surjective (@id X) :=
assume y,
show  x, id x = y, from exists.intro y rfl

theorem bijective_id : bijective (@id X) :=
and.intro injective_id surjective_id

More interestingly, we can prove that the composition of injective functions is injective, and so on.

theorem injective_comp {g : Y  Z} {f : X  Y}
    (Hg : injective g) (Hf : injective f) :
  injective (g  f) :=
assume x₁ x₂,
assume : (g  f) x₁ = (g  f) x₂,
have f x₁ = f x₂, from Hg this,
show x₁ = x₂, from Hf this

theorem surjective_comp {g : Y  Z} {f : X  Y}
    (hg : surjective g) (hf : surjective f) :
  surjective (g  f) :=
assume z,
exists.elim (hg z) $
assume y (hy : g y = z),
exists.elim (hf y) $
assume x (hx : f x = y),
have g (f x) = z, from eq.subst (eq.symm hx) hy,
show  x, g (f x) = z, from exists.intro x this

theorem bijective_comp {g : Y  Z} {f : X  Y}
    (hg : bijective g) (hf : bijective f) :
  bijective (g  f) :=
have ginj : injective g, from hg.left,
have gsurj : surjective g, from hg.right,
have finj : injective f, from hf.left,
have fsurj : surjective f, from hf.right,
and.intro (injective_comp ginj finj)
  (surjective_comp gsurj fsurj)

The notions of left and right inverse are defined in the expected way.

-- g is a left inverse to f
def left_inverse (g : Y  X) (f : X  Y) : Prop :=  x, g (f x) = x

-- g is a right inverse to f
def right_inverse (g : Y  X) (f : X  Y) : Prop := left_inverse f g

In particular, composing with a left or right inverse yields the identity.

def id_of_left_inverse {g : Y  X} {f : X  Y} : left_inverse g f  g  f = id :=
assume H, funext H

def id_of_right_inverse {g : Y  X} {f : X  Y} : right_inverse g f  f  g = id :=
assume H, funext H

Notice that we need to use funext to show the equality of functions.

The following shows that if a function has a left inverse, then it is injective, and if it has a right inverse, then it is surjective.

theorem injective_of_left_inverse {g : Y  X} {f : X  Y} :
  left_inverse g f  injective f :=
assume h, assume x₁ x₂, assume feq,
calc x₁ = g (f x₁) : by rw h
    ... = g (f x₂) : by rw feq
    ... = x₂       : by rw h

theorem surjective_of_right_inverse {g : Y   X} {f : X  Y} :
  right_inverse g f  surjective f :=
assume h, assume y,
let  x : X := g y in
have f x = y, from calc
  f x  = (f (g y))    : rfl
   ... = y            : by rw [h y],
show  x, f x = y, from exists.intro x this

16.4. Defining the Inverse Classically

All the theorems listed in the previous section are found in the Lean library, and are available to you when you open the function namespace with open function:

open function

#check comp
#check left_inverse
#check has_right_inverse

Defining inverse functions, however, requires classical reasoning, which we get by opening the classical namespace:

open classical

section
  variables A B : Type
  variable P : A  Prop
  variable R : A  B  Prop

  example : ( x,  y, R x y)   f : A  B,  x, R x (f x) :=
  axiom_of_choice

  example (h :  x, P x) : P (some h) :=
  some_spec h
end

The axiom of choice tells us that if, for every x : X, there is a y : Y satisfying R x y, then there is a function f : X Y which, for every x chooses such a y. In Lean, this “axiom” is proved using a classical construction, the some function (sometimes called “the indefinite description operator”) which, given that there is some x satisfying P x, returns such an x. With these constructions, the inverse function is defined as follows:

open classical function
local attribute [instance] prop_decidable

variables {X Y : Type}

noncomputable def inverse (f : X  Y) (default : X) : Y  X :=
λ y, if h :  x, f x = y then some h else default

Lean requires us to acknowledge that the definition is not computational, since, first, it may not be algorithmically possible to decide whether or not condition H holds, and even if it does, it may not be algorithmically possible to find a suitable value of x.

Below, the proposition inverse_of_exists asserts that inverse meets its specification, and the subsequent theorem shows that if f is injective, then the inverse function really is a left inverse.

theorem inverse_of_exists (f : X  Y) (default : X) (y : Y)
  (h :  x, f x = y) :
f (inverse f default y) = y :=
have h1 : inverse f default y = some h, from dif_pos h,
have h2 : f (some h) = y, from some_spec h,
eq.subst (eq.symm h1) h2

theorem is_left_inverse_of_injective (f : X  Y) (default : X)
  (injf : injective f) :
left_inverse (inverse f default) f :=
let finv := (inverse f default) in
assume x,
have h1 :  x', f x' = f x, from exists.intro x rfl,
have h2 : f (finv (f x)) = f x, from inverse_of_exists f default (f x) h1,
show finv (f x) = x, from injf h2

16.5. Functions and Sets in Lean

In Section 7.4 we saw how to represent relativized universal and existential quantifiers when formalizing phrases like “every prime number greater than two is odd” and “some prime number is even.” In a similar way, we can relativize statements to sets. In symbolic logic, the expression \(\exists x \in A \; P (x)\) abbreviates \(\exists x \; (x \in A \wedge P(x))\), and \(\forall x \in A \; P (x)\) abbreviates \(\forall x \; (x \in A \to P(x))\).

Lean also defines notation for relativized quantifiers:

variables (X : Type) (A : set X) (P : X  Prop)

#check  x  A, P x
#check  x  A, P x

Here is an example of how to use the bounded universal quantifier:

example (h :  x  A, P x) (x : X) (h1 : x  A) : P x := h x h1

Using bounded quantifiers, we can talk about the behavior of functions on particular sets:

import data.set
open set function

variables {X Y : Type}
variables (A  : set X) (B : set Y)

def maps_to (f : X  Y) (A : set X) (B : set Y) :=  x  A, f x  B

def inj_on (f : X  Y) (A : set X) :=  (x₁  A) (x₂  A), f x₁ = f x₂  x₁ = x₂

def surj_on (f : X  Y) (A : set X) (B : set Y) := B  f '' A

The expression maps_to f A B asserts that f maps elements of the set A to the set B, and the expression inj_on f A asserts that f is injective on A. The expression surj_on f A B asserts that, viewed as a function defined on elements of A, the function f is surjective onto the set B. Here are examples of how they can be used:

variables (f : X  Y) (A : set X) (B : set Y)

example (h : maps_to f A B) (x : X) (h1 : x  A) : f x  B := h x h1

example (h : inj_on f A) (x₁ x₂ : X) (h1 : x₁  A) (h2 : x₂  A)
    (h3 : f x₁ = f x₂) : x₁ = x₂ :=
h x₁ h1 x₂ h2 h3

example (h : surj_on f A B) (y : Y) (h1 : y  B) :  x, x  A  f x = y :=
h h1

Actually, it is slightly more convenient to mark the variables in maps_to and inj_on as implicit:

def maps_to (f : X  Y) (A : set X) (B : set Y) :=  {x}, x  A  f x  B

def inj_on (f : X  Y) (A : set X) :=  {x₁ x₂}, x₁  A  x₂  A  f x₁ = f x₂  x₁ = x₂

In that case, we can leave out some arguments:

variables (f : X  Y) (A : set X) (B : set Y)

example (h : maps_to f A B) (x : X) (h1 : x  A) : f x  B := h h1

example (h : inj_on f A) (x₁ x₂ : X) (h1 : x₁  A) (h2 : x₂  A)
    (h3 : f x₁ = f x₂) : x₁ = x₂ :=
h h1 h2 h3

In the examples below, we’ll use the versions with implicit arguments.

The expression surj_on f A B asserts that, viewed as a function defined on elements of A, the function f is surjective onto the set B:

With these notions in hand, we can prove that the composition of injective functions is injective. The proof is similar to the one above, though now we have to be more careful to relativize claims to A and B:

theorem inj_on_comp (fAB : maps_to f A B) (hg : inj_on g B) (hf: inj_on f A) :
  inj_on (g  f) A :=
assume x1 x2 : X,
assume x1A : x1  A,
assume x2A : x2  A,
have fx1B : f x1  B, from fAB x1A,
have fx2B : f x2  B, from fAB x2A,
assume h1 : g (f x1) = g (f x2),
have h2 : f x1 = f x2, from hg fx1B fx2B h1,
show x1 = x2, from hf x1A x2A h2

We can similarly prove that the composition of surjective functions is surjective:

theorem surj_on_comp (hg : surj_on g B C) (hf: surj_on f A B) :
  surj_on (g  f) A C :=
assume z,
assume zc : z  C,
exists.elim (hg zc) $
assume y (h1 : y  B  g y = z),
exists.elim (hf (and.left h1)) $
assume x (h2 : x  A  f x = y),
show x, x  A  g (f x) = z, from
  exists.intro x
    (and.intro
      (and.left h2)
      (calc
        g (f x) = g y : by rw and.right h2
            ... = z   : by rw and.right h1))

The following shows that the image of a union is the union of images:

theorem image_union : f '' (A₁  A₂) =f '' A₁  f '' A₂ :=
ext (assume y, iff.intro
  (assume h : y  image f (A₁  A₂),
    exists.elim h $
    assume x h1,
    have xA₁A₂ : x  A₁  A₂, from h1.left,
    have fxy : f x = y, from h1.right,
    or.elim xA₁A₂
      (assume xA₁, or.inl (mem_image xA₁ fxy))
      (assume xA₂, or.inr (mem_image xA₂ fxy)))
  (assume h : y  image f A₁  image f A₂,
    or.elim h
      (assume yifA₁ : y  image f A₁,
        exists.elim yifA₁ $
        assume x h1,
        have xA₁ : x  A₁, from h1.left,
        have fxy : f x = y, from h1.right,
        mem_image (or.inl xA₁) fxy)
      (assume yifA₂ : y  image f A₂,
        exists.elim yifA₂ $
        assume x h1,
        have xA₂ : x  A₂, from h1.left,
        have fxy : f x = y, from h1.right,
        mem_image (or.inr xA₂) fxy)))

16.6. Exercises

  1. Fill in the sorry’s in the last three proofs below.

    open function int algebra
    
    def f (x : ) :  := x + 3
    def g (x : ) :  := -x
    def h (x : ) :  := 2 * x + 3
    
    example : injective f :=
    assume x1 x2,
    assume h1 : x1 + 3 = x2 + 3,   -- Lean knows this is the same as f x1 = f x2
    show x1 = x2, from eq_of_add_eq_add_right h1
    
    example : surjective f :=
    assume y,
    have h1 : f (y - 3) = y, from calc
      f (y - 3) = (y - 3) + 3 : rfl
            ... = y           : by rw sub_add_cancel,
    show  x, f x = y, from exists.intro (y - 3) h1
    
    example (x y : ) (h : 2 * x = 2 * y) : x = y :=
    have h1 : 2  (0 : ), from dec_trivial,  -- this tells Lean to figure it out itself
    show x = y, from eq_of_mul_eq_mul_left h1 h
    
    example (x : ) : -(-x) = x := neg_neg x
    
    example (A B : Type) (u : A  B) (v : B  A) (h : left_inverse u v) :
       x, u (v x) = x :=
    h
    
    example (A B : Type) (u : A  B) (v : B  A) (h : left_inverse u v) :
      right_inverse v u :=
    h
    
    -- fill in the sorry's in the following proofs
    
    example : injective h :=
    sorry
    
    example : surjective g :=
    sorry
    
    example (A B : Type) (u : A  B) (v1 : B  A) (v2 : B  A)
      (h1 : left_inverse v1 u) (h2 : right_inverse v2 u) : v1 = v2 :=
    funext
      (assume x,
        calc
          v1 x = v1 (u (v2 x)) : sorry
           ... = v2 x          : sorry)
    
  2. Fill in the sorry in the proof below.

    import data.set
    open function set
    
    variables {X Y : Type}
    variable  f : X  Y
    variables A B : set X
    
    example : f '' (A  B) = f '' A  f '' B :=
    eq_of_subset_of_subset
      (assume y,
        assume h1 : y  f '' (A  B),
        exists.elim h1 $
        assume x h,
        have h2 : x  A  B, from h.left,
        have h3 : f x = y, from h.right,
        or.elim h2
          (assume h4 : x  A,
            have h5 : y  f '' A, from mem_image h4 h3,
            show y  f '' A  f '' B, from or.inl h5)
          (assume h4 : x  B,
            have h5 : y  f ''  B, from mem_image h4 h3,
            show y  f '' A  f '' B, from or.inr h5))
      (assume y,
        assume h2 : y  f '' A  f '' B,
        or.elim h2
          (assume h3 : y  f '' A,
            exists.elim h3 $
            assume x h,
            have h4 : x  A, from h.left,
            have h5 : f x = y, from h.right,
            have h6 : x  A  B, from or.inl h4,
            show y  f '' (A  B), from mem_image h6 h5)
          (assume h3 : y  f '' B,
            exists.elim h3 $
            assume x h,
            have h4 : x  B, from h.left,
            have h5 : f x = y, from h.right,
            have h6 : x  A  B, from or.inr h4,
            show y  f '' (A  B), from mem_image h6 h5))
    
    -- remember, x ∈ A ∩ B is the same as x ∈ A ∧ x ∈ B
    example (x : X) (h1 : x  A) (h2 : x  B) : x  A  B :=
    and.intro h1 h2
    
    example (x : X) (h1 : x  A  B) : x  A :=
    and.left h1
    
    -- Fill in the proof below.
    -- (It should take about 8 lines.)
    
    example : f '' (A  B)  f '' A  f '' B :=
    assume y,
    assume h1 : y  f '' (A  B),
    show y  f '' A  f '' B, from sorry