# 14. Relations in Lean¶

In the last chapter, we noted that set theorists think of a binary relation \(R\) on a set \(A\) as a set of ordered pairs, so that \(R(a, b)\) really means \((a, b) \in R\). An alternative is to think of \(R\) as a function which, when applied to \(a\) and \(B\), returns the proposition that \(R(a, b)\) holds. This is the viewpoint adopted by Lean: a binary relation on a type `A`

is a function `A → A → Prop`

. Remember that the arrows associate to the right, so `A → A → Prop`

really means `A → (A → Prop)`

. So, given `a : A`

, `R a`

is a predicate (the property of being related to `A`

), and given `a b : A`

, `R a b`

is a proposition.

## 14.1. Order Relations¶

With first-order logic, we can say what it means for a relation to be reflexive, symmetric, transitive, antisymmetric, and so on:

```
namespace hidden
variable {A : Type}
def reflexive (R : A → A → Prop) : Prop :=
∀ x, R x x
def symmetric (R : A → A → Prop) : Prop :=
∀ x y, R x y → R y x
def transitive (R : A → A → Prop) : Prop :=
∀ x y z, R x y → R y z → R x z
def antisymmetric (R : A → A → Prop) : Prop :=
∀ x y, R x y → R y x → x = y
end hidden
```

We can then use the notions freely. Notice that Lean will unfold the definitions when necessary, for example, treating `reflexive R`

as `∀ x, R x x`

.

```
variable R : A → A → Prop
example (h : reflexive R) (x : A) : R x x := h x
example (h : symmetric R) (x y : A) (h1 : R x y) : R y x :=
h x y h1
example (h : transitive R) (x y z : A) (h1 : R x y) (h2 : R y z) :
R x z :=
h x y z h1 h2
example (h : antisymmetric R) (x y : A) (h1 : R x y)
(h2 : R y x) :
x = y :=
h x y h1 h2
```

In the command `variable {A : Type}`

, we put curly braces around `A`

to indicate that it is an *implicit* argument, which is to say, you do not have to write it explicitly; Lean can infer it from the argument `R`

. That is why we can write `reflexive R`

rather than `reflexive A R`

: Lean knows that `R`

is a binary relation on `A`

, so it can infer that we mean reflexivity for binary relations on `A`

.

Given `h : transitive R`

, `h1 : R x y`

, and `h2 : R y z`

, it is annoying to have to write `h x y z h1 h2`

to prove `R x z`

. After all, Lean should be able to infer that we are talking about transitivity at `x`

, `y`

, and `z`

, from the fact that `h1`

is `R x y`

and `h2`

is `R y z`

. Indeed, we can replace that information by underscores:

```
variable R : A → A → Prop
example (h : transitive R) (x y z : A) (h1 : R x y)
(h2 : R y z) :
R x z :=
h _ _ _ h1 h2
```

But typing underscores is annoying, too. The best solution is to declare the arguments `x y z`

to a transitivity hypothesis to be implicit as well:

```
variable {A : Type}
def transitive (R : A → A → Prop) : Prop :=
∀ {x y z}, R x y → R y z → R x z
variable R : A → A → Prop
example (h : transitive R) (x y z : A) (h1 : R x y) (h2 : R y z) :
R x z :=
h h1 h2
```

In fact, the notions `reflexive`

, `symmetric`

, `transitive`

, and `antisymmetric`

are defined in Lean’s core library in exactly this way, so we are free to use them without defining them. (That is why we put our temporary definitions of in a namespace `hidden`

; that means that the full name of our version of `reflexive`

is `hidden.reflexive`

, which, therefore, doesn’t conflict with the one defined in the library.)

In Section 13.1 we showed that a strict partial order—that is, a binary relation that is transitive and irreflexive—is also asymmetric. Here is a proof of that fact in Lean.

```
variable A : Type
variable R : A → A → Prop
example (h1 : irreflexive R) (h2 : transitive R) :
∀ x y, R x y → ¬ R y x :=
assume x y,
assume h3 : R x y,
assume h4 : R y x,
have h5 : R x x, from h2 h3 h4,
have h6 : ¬ R x x, from h1 x,
show false, from h6 h5
```

In mathematics, it is common to use infix notation and a symbol like `≤`

to denote a partial order. Lean supports this practice:

```
section
parameter A : Type
parameter R : A → A → Prop
local infix ≤ := R
example (h1 : irreflexive R) (h2 : transitive R) :
∀ x y, x ≤ y → ¬ y ≤ x :=
assume x y,
assume h3 : x ≤ y,
assume h4 : y ≤ x,
have h5 : x ≤ x, from h2 h3 h4,
have h6 : ¬ x ≤ x, from h1 x,
show false, from h6 h5
end
```

The `parameter`

and `parameters`

commands are similar to the `variable`

and `variables`

commands, except that parameters are fixed within a section. In other words, if you prove a theorem about `R`

in the section above, you cannot apply that theorem to another relation, `S`

, without closing the section. Since the parameter `R`

is fixed, Lean allows us to define notation for `R`

to be used locally in the section.

In the example below, having fixed a partial order, `R`

, we define the corresponding strict partial order and prove that it is, indeed, a strict order.

```
section
parameters {A : Type} (R : A → A → Prop)
parameter (reflR : reflexive R)
parameter (transR : transitive R)
parameter (antisymmR : ∀ {a b : A}, R a b → R b a → a = b)
local infix ≤ := R
definition R' (a b : A) : Prop := a ≤ b ∧ a ≠ b
local infix < := R'
theorem irreflR (a : A) : ¬ a < a :=
assume : a < a,
have a ≠ a, from and.right this,
have a = a, from rfl,
show false, from ‹a ≠ a› ‹a = a›
theorem transR {a b c : A} (h₁ : a < b) (h₂ : b < c) : a < c :=
have a ≤ b, from and.left h₁,
have a ≠ b, from and.right h₁,
have b ≤ c, from and.left h₂,
have b ≠ c, from and.right h₂,
have a ≤ c, from transR ‹a ≤ b› ‹b ≤ c›,
have a ≠ c, from
assume : a = c,
have c ≤ b, from eq.subst ‹a = c› ‹a ≤ b›,
have b = c, from antisymmR ‹b ≤ c› ‹c ≤ b›,
show false, from ‹b ≠ c› ‹b = c›,
show a < c, from and.intro ‹a ≤ c› ‹a ≠ c›
end
```

Notice that we have used suggestive names `reflR`

, `transR`

, `antisymmR`

instead of `h1`

, `h2`

, `h3`

to help remember which hypothesis is which. The proof also uses anonymous `have`

and `assume`

, referring back to them with the French quotes, `\f<`

anf `\f>`

. Remember also that `eq.subst ‹a = c› ‹a ≤ b›`

is a proof of the fact that amounts for substituting `c`

for `a`

in `a ≤ b`

. You can also use the equivalent notation `‹a = c› ▸ ‹a ≤ b›`

, where the triangle is written `\t`

.

In Section Section 13.1, we also noted that you can define a (weak) partial order from a strict one. We ask you to do this formally in the exercises below.

Here is one more example. Suppose `R`

is a binary relation on a type `A`

, and we define `S x y`

to mean that both `R x y`

and `R y x`

holds. Below we show that the resulting relation is reflexive and symmetric.

```
section
parameter A : Type
parameter R : A → A → Prop
variable h1 : transitive R
variable h2 : reflexive R
def S (x y : A) := R x y ∧ R y x
example : reflexive S :=
assume x,
have R x x, from h2 x,
show S x x, from and.intro this this
example : symmetric S :=
assume x y,
assume h : S x y,
have h1 : R x y, from h.left,
have h2 : R y x, from h.right,
show S y x, from ⟨h.right, h.left⟩
end
```

In the exercises below, we ask you to show that `S`

is transitive as well.

In the first example, we use the anonymous `assume`

and `have`

, and then refer back to the `have`

with the keyword `this`

. In the second example, we abbreviate `and.left h`

and `and.right h`

as `h.left`

and `h.right`

, respectively. We also abbreviate `and.intro h.right h.left`

with an anonymous constructor, writing `⟨h.right, h.left⟩`

. Lean figures out that we are trying to prove a conjunction, and figures out that `and.intro`

is the relevant introduction principle. You can type the corner brackets with `\<`

and `\>`

, respectively.

## 14.2. Orderings on Numbers¶

Conveniently, Lean has the normal orderings on the natural numbers, integers, and so on defined already.

```
open nat
variables n m : ℕ
#check 0 ≤ n
#check n < n + 1
example : 0 ≤ n := zero_le n
example : n < n + 1 := lt_succ_self n
example (h : n + 1 ≤ m) : n < m + 1 :=
have h1 : n < n + 1, from lt_succ_self n,
have h2 : n < m, from lt_of_lt_of_le h1 h,
have h3 : m < m + 1, from lt_succ_self m,
show n < m + 1, from lt.trans h2 h3
```

There are many theorems in Lean that are useful for proving facts about inequality relations. We list some common ones here.

```
variables (A : Type) [partial_order A]
variables a b c : A
#check (le_trans : a ≤ b → b ≤ c → a ≤ c)
#check (lt_trans : a < b → b < c → a < c)
#check (lt_of_lt_of_le : a < b → b ≤ c → a < c)
#check (lt_of_le_of_lt : a ≤ b → b < c → a < c)
#check (le_of_lt : a < b → a ≤ b)
```

Here the declaration at the top says that `A`

has the structure of a partial order. There are also properties that are specific to some domains, like the natural numbers:

```
variable n : ℕ
#check (nat.zero_le : ∀ n : ℕ, 0 ≤ n)
#check (nat.lt_succ_self : ∀ n : ℕ, n < n + 1)
#check (nat.le_succ : ∀ n : ℕ, n ≤ n + 1)
```

## 14.3. Exercises¶

Replace the

`sorry`

commands in the following proofs to show that we can create a partial order`R'`

out of a strict partial order`R`

.section parameters {A : Type} {R : A → A → Prop} parameter (irreflR : irreflexive R) parameter (transR : transitive R) local infix < := R def R' (a b : A) : Prop := R a b ∨ a = b local infix ≤ := R' theorem reflR' (a : A) : a ≤ a := sorry theorem transR' {a b c : A} (h1 : a ≤ b) (h2 : b ≤ c): a ≤ c := sorry theorem antisymmR' {a b : A} (h1 : a ≤ b) (h2 : b ≤ a) : a = b := sorry end

Replace the

`sorry`

by a proof.section parameters {A : Type} {R : A → A → Prop} parameter (reflR : reflexive R) parameter (transR : transitive R) def S (a b : A) : Prop := R a b ∧ R b a example : transitive S := sorry end

Only one of the following two theorems is provable. Figure out which one is true, and replace the

`sorry`

command with a complete proof.section parameters {A : Type} {a b c : A} {R : A → A → Prop} parameter (Rab : R a b) parameter (Rbc : R b c) parameter (nRac : ¬ R a c) -- Prove one of the following two theorems: theorem R_is_strict_partial_order : irreflexive R ∧ transitive R := sorry theorem R_is_not_strict_partial_order : ¬(irreflexive R ∧ transitive R) := sorry end

Complete the following proof.

open nat example : 1 ≤ 4 := sorry