# 3. Logic

In the last chapter, we dealt with equations, inequalities, and basic mathematical statements like “$$x$$ divides $$y$$.” Complex mathematical statements are built up from simple ones like these using logical terms like “and,” “or,” “not,” and “if … then,” “every,” and “some.” In this chapter, we show you how to work with statements that are built up in this way.

## 3.1. Implication and the Universal Quantifier

Consider the statement after the #check:

#check ∀ x : ℝ, 0 ≤ x → |x| = x


In words, we would say “for every real number x, if 0 ≤ x then the absolute value of x equals x”. We can also have more complicated statements like:

#check ∀ x y ε : ℝ, 0 < ε → ε ≤ 1 → |x| < ε → |y| < ε → |x * y| < ε


In words, we would say “for every x, y, and ε, if 0 < ε ≤ 1, the absolute value of x is less than ε, and the absolute value of y is less than ε, then the absolute value of x * y is less than ε.” In Lean, in a sequence of implications there are implicit parentheses grouped to the right. So the expression above means “if 0 < ε then if ε ≤ 1 then if |x| < ε …” As a result, the expression says that all the assumptions together imply the conclusion.

You have already seen that even though the universal quantifier in this statement ranges over objects and the implication arrows introduce hypotheses, Lean treats the two in very similar ways. In particular, if you have proved a theorem of that form, you can apply it to objects and hypotheses in the same way. We will use as an example the following statement that we will help you to prove a bit later:

theorem my_lemma : ∀ x y ε : ℝ, 0 < ε → ε ≤ 1 → |x| < ε → |y| < ε → |x * y| < ε :=
sorry

section
variable (a b δ : ℝ)
variable (h₀ : 0 < δ) (h₁ : δ ≤ 1)
variable (ha : |a| < δ) (hb : |b| < δ)

#check my_lemma a b δ
#check my_lemma a b δ h₀ h₁
#check my_lemma a b δ h₀ h₁ ha hb

end


You have also already seen that it is common in Lean to use curly brackets to make quantified variables implicit when they can be inferred from subsequent hypotheses. When we do that, we can just apply a lemma to the hypotheses without mentioning the objects.

theorem my_lemma2 : ∀ {x y ε : ℝ}, 0 < ε → ε ≤ 1 → |x| < ε → |y| < ε → |x * y| < ε :=
sorry

section
variable (a b δ : ℝ)
variable (h₀ : 0 < δ) (h₁ : δ ≤ 1)
variable (ha : |a| < δ) (hb : |b| < δ)

#check my_lemma2 h₀ h₁ ha hb

end


At this stage, you also know that if you use the apply tactic to apply my_lemma to a goal of the form |a * b| < δ, you are left with new goals that require you to prove each of the hypotheses.

To prove a statement like this, use the intro tactic. Take a look at what it does in this example:

theorem my_lemma3 :
∀ {x y ε : ℝ}, 0 < ε → ε ≤ 1 → |x| < ε → |y| < ε → |x * y| < ε := by
intro x y ε epos ele1 xlt ylt
sorry


We can use any names we want for the universally quantified variables; they do not have to be x, y, and ε. Notice that we have to introduce the variables even though they are marked implicit: making them implicit means that we leave them out when we write an expression using my_lemma, but they are still an essential part of the statement that we are proving. After the intro command, the goal is what it would have been at the start if we listed all the variables and hypotheses before the colon, as we did in the last section. In a moment, we will see why it is sometimes necessary to introduce variables and hypotheses after the proof begins.

theorem my_lemma4 :
∀ {x y ε : ℝ}, 0 < ε → ε ≤ 1 → |x| < ε → |y| < ε → |x * y| < ε := by
intro x y ε epos ele1 xlt ylt
calc
|x * y| = |x| * |y| := sorry
_ ≤ |x| * ε := sorry
_ < 1 * ε := sorry
_ = ε := sorry


Finish the proof using the theorems abs_mul, mul_le_mul, abs_nonneg, mul_lt_mul_right, and one_mul. Remember that you can find theorems like these using Ctrl-space completion (or Cmd-space completion on a Mac). Remember also that you can use .mp and .mpr or .1 and .2 to extract the two directions of an if-and-only-if statement.

Universal quantifiers are often hidden in definitions, and Lean will unfold definitions to expose them when necessary. For example, let’s define two predicates, FnUb f a and FnLb f a, where f is a function from the real numbers to the real numbers and a is a real number. The first says that a is an upper bound on the values of f, and the second says that a is a lower bound on the values of f.

def FnUb (f : ℝ → ℝ) (a : ℝ) : Prop :=
∀ x, f x ≤ a

def FnLb (f : ℝ → ℝ) (a : ℝ) : Prop :=
∀ x, a ≤ f x


In the next example, fun x ↦ f x + g x is the function that maps x to f x + g x. Going from the expression f x + g x to this function is called a lambda abstraction in type theory.

example (hfa : FnUb f a) (hgb : FnUb g b) : FnUb (fun x ↦ f x + g x) (a + b) := by
intro x
dsimp
apply hfa
apply hgb


Applying intro to the goal FnUb (fun x ↦ f x + g x) (a + b) forces Lean to unfold the definition of FnUb and introduce x for the universal quantifier. The goal is then (fun (x : ℝ) ↦ f x + g x) x ≤ a + b. But applying (fun x ↦ f x + g x) to x should result in f x + g x, and the dsimp command performs that simplification. (The “d” stands for “definitional.”) You can delete that command and the proof still works; Lean would have to perform that contraction anyhow to make sense of the next apply. The dsimp command simply makes the goal more readable and helps us figure out what to do next. Another option is to use the change tactic by writing change f x + g x ≤ a + b. This helps make the proof more readable, and gives you more control over how the goal is transformed.

The rest of the proof is routine. The last two apply commands force Lean to unfold the definitions of FnUb in the hypotheses. Try carrying out similar proofs of these:

example (hfa : FnLb f a) (hgb : FnLb g b) : FnLb (fun x ↦ f x + g x) (a + b) :=
sorry

example (nnf : FnLb f 0) (nng : FnLb g 0) : FnLb (fun x ↦ f x * g x) 0 :=
sorry

example (hfa : FnUb f a) (hgb : FnUb g b) (nng : FnLb g 0) (nna : 0 ≤ a) :
FnUb (fun x ↦ f x * g x) (a * b) :=
sorry


Even though we have defined FnUb and FnLb for functions from the reals to the reals, you should recognize that the definitions and proofs are much more general. The definitions make sense for functions between any two types for which there is a notion of order on the codomain. Checking the type of the theorem add_le_add shows that it holds of any structure that is an “ordered additive commutative monoid”; the details of what that means don’t matter now, but it is worth knowing that the natural numbers, integers, rationals, and real numbers are all instances. So if we prove the theorem fnUb_add at that level of generality, it will apply in all these instances.

variable {α : Type*} {R : Type*} [OrderedCancelAddCommMonoid R]

def FnUb' (f : α → R) (a : R) : Prop :=
∀ x, f x ≤ a

theorem fnUb_add {f g : α → R} {a b : R} (hfa : FnUb' f a) (hgb : FnUb' g b) :
FnUb' (fun x ↦ f x + g x) (a + b) := fun x ↦ add_le_add (hfa x) (hgb x)


You have already seen square brackets like these in Section Section 2.2, though we still haven’t explained what they mean. For concreteness, we will stick to the real numbers for most of our examples, but it is worth knowing that Mathlib contains definitions and theorems that work at a high level of generality.

For another example of a hidden universal quantifier, Mathlib defines a predicate Monotone, which says that a function is nondecreasing in its arguments:

example (f : ℝ → ℝ) (h : Monotone f) : ∀ {a b}, a ≤ b → f a ≤ f b :=
@h


The property Monotone f is defined to be exactly the expression after the colon. We need to put the @ symbol before h because if we don’t, Lean expands the implicit arguments to h and inserts placeholders.

Proving statements about monotonicity involves using intro to introduce two variables, say, a and b, and the hypothesis a ≤ b. To use a monotonicity hypothesis, you can apply it to suitable arguments and hypotheses, and then apply the resulting expression to the goal. Or you can apply it to the goal and let Lean help you work backwards by displaying the remaining hypotheses as new subgoals.

example (mf : Monotone f) (mg : Monotone g) : Monotone fun x ↦ f x + g x := by
intro a b aleb
apply mf aleb
apply mg aleb


When a proof is this short, it is often convenient to give a proof term instead. To describe a proof that temporarily introduces objects a and b and a hypothesis aleb, Lean uses the notation fun a b aleb ↦ .... This is analogous to the way that an expression like fun x ↦ x^2 describes a function by temporarily naming an object, x, and then using it to describe a value. So the intro command in the previous proof corresponds to the lambda abstraction in the next proof term. The apply commands then correspond to building the application of the theorem to its arguments.

example (mf : Monotone f) (mg : Monotone g) : Monotone fun x ↦ f x + g x :=


Here is a useful trick: if you start writing the proof term fun a b aleb ↦ _ using an underscore where the rest of the expression should go, Lean will flag an error, indicating that it can’t guess the value of that expression. If you check the Lean Goal window in VS Code or hover over the squiggly error marker, Lean will show you the goal that the remaining expression has to solve.

Try proving these, with either tactics or proof terms:

example {c : ℝ} (mf : Monotone f) (nnc : 0 ≤ c) : Monotone fun x ↦ c * f x :=
sorry

example (mf : Monotone f) (mg : Monotone g) : Monotone fun x ↦ f (g x) :=
sorry


Here are some more examples. A function $$f$$ from $$\Bbb R$$ to $$\Bbb R$$ is said to be even if $$f(-x) = f(x)$$ for every $$x$$, and odd if $$f(-x) = -f(x)$$ for every $$x$$. The following example defines these two notions formally and establishes one fact about them. You can complete the proofs of the others.

def FnEven (f : ℝ → ℝ) : Prop :=
∀ x, f x = f (-x)

def FnOdd (f : ℝ → ℝ) : Prop :=
∀ x, f x = -f (-x)

example (ef : FnEven f) (eg : FnEven g) : FnEven fun x ↦ f x + g x := by
intro x
calc
(fun x ↦ f x + g x) x = f x + g x := rfl
_ = f (-x) + g (-x) := by rw [ef, eg]

example (of : FnOdd f) (og : FnOdd g) : FnEven fun x ↦ f x * g x := by
sorry

example (ef : FnEven f) (og : FnOdd g) : FnOdd fun x ↦ f x * g x := by
sorry

example (ef : FnEven f) (og : FnOdd g) : FnEven fun x ↦ f (g x) := by
sorry


The first proof can be shortened using dsimp or change to get rid of the lambda abstraction. But you can check that the subsequent rw won’t work unless we get rid of the lambda abstraction explicitly, because otherwise it cannot find the patterns f x and g x in the expression. Contrary to some other tactics, rw operates on the syntactic level, it won’t unfold definitions or apply reductions for you (it has a variant called erw that tries a little harder in this direction, but not much harder).

You can find implicit universal quantifiers all over the place, once you know how to spot them.

Mathlib includes a good library for manipulating sets. Recall that Lean does not use foundations based on set theory, so here the word set has its mundane meaning of a collection of mathematical objects of some given type α. If x has type α and s has type Set α, then x ∈ s is a proposition that asserts that x is an element of s. If y has some different type β then the expression y ∈ s makes no sense. Here “makes no sense” means “has no type hence Lean does not accept it as a well-formed statement”. This contrasts with Zermelo-Fraenkel set theory for instance where a ∈ b is a well-formed statement for every mathematical objects a and b. For instance sin ∈ cos is a well-formed statement in ZF. This defect of set theoretic foundations is an important motivation for not using it in a proof assistant which is meant to assist us by detecting meaningless expressions. In Lean sin has type ℝ → ℝ and cos has type ℝ → ℝ which is not equal to Set (ℝ → ℝ), even after unfolding definitions, so the statement sin ∈ cos makes no sense. One can also use Lean to work on set theory itself. For instance the independence of the continuum hypothesis from the axioms of Zermelo-Fraenkel has been formalized in Lean. But such a meta-theory of set theory is completely beyond the scope of this book.

If s and t are of type Set α, then the subset relation s ⊆ t is defined to mean ∀ {x : α}, x ∈ s → x ∈ t. The variable in the quantifier is marked implicit so that given h : s ⊆ t and h' : x ∈ s, we can write h h' as justification for x ∈ t. The following example provides a tactic proof and a proof term justifying the reflexivity of the subset relation, and asks you to do the same for transitivity.

variable {α : Type*} (r s t : Set α)

example : s ⊆ s := by
intro x xs
exact xs

theorem Subset.refl : s ⊆ s := fun x xs ↦ xs

theorem Subset.trans : r ⊆ s → s ⊆ t → r ⊆ t := by
sorry


Just as we defined FnUb for functions, we can define SetUb s a to mean that a is an upper bound on the set s, assuming s is a set of elements of some type that has an order associated with it. In the next example, we ask you to prove that if a is a bound on s and a ≤ b, then b is a bound on s as well.

variable {α : Type*} [PartialOrder α]
variable (s : Set α) (a b : α)

def SetUb (s : Set α) (a : α) :=
∀ x, x ∈ s → x ≤ a

example (h : SetUb s a) (h' : a ≤ b) : SetUb s b :=
sorry


We close this section with one last important example. A function $$f$$ is said to be injective if for every $$x_1$$ and $$x_2$$, if $$f(x_1) = f(x_2)$$ then $$x_1 = x_2$$. Mathlib defines Function.Injective f with x₁ and x₂ implicit. The next example shows that, on the real numbers, any function that adds a constant is injective. We then ask you to show that multiplication by a nonzero constant is also injective, using the lemma name in the example as a source of inspiration. Recall you should use Ctrl-space completion after guessing the beginning of a lemma name.

open Function

example (c : ℝ) : Injective fun x ↦ x + c := by
intro x₁ x₂ h'

example {c : ℝ} (h : c ≠ 0) : Injective fun x ↦ c * x := by
sorry


Finally, show that the composition of two injective functions is injective:

variable {α : Type*} {β : Type*} {γ : Type*}
variable {g : β → γ} {f : α → β}

example (injg : Injective g) (injf : Injective f) : Injective fun x ↦ g (f x) := by
sorry


## 3.2. The Existential Quantifier

The existential quantifier, which can be entered as \ex in VS Code, is used to represent the phrase “there exists.” The formal expression ∃ x : ℝ, 2 < x ∧ x < 3 in Lean says that there is a real number between 2 and 3. (We will discuss the conjunction symbol, ∧, in Section 3.4.) The canonical way to prove such a statement is to exhibit a real number and show that it has the stated property. The number 2.5, which we can enter as 5 / 2 or (5 : ℝ) / 2 when Lean cannot infer from context that we have the real numbers in mind, has the required property, and the norm_num tactic can prove that it meets the description.

There are a few ways we can put the information together. Given a goal that begins with an existential quantifier, the use tactic is used to provide the object, leaving the goal of proving the property.

example : ∃ x : ℝ, 2 < x ∧ x < 3 := by
use 5 / 2
norm_num


You can give the use tactic proofs as well as data:

example : ∃ x : ℝ, 2 < x ∧ x < 3 := by
have h1 : 2 < (5 : ℝ) / 2 := by norm_num
have h2 : (5 : ℝ) / 2 < 3 := by norm_num
use 5 / 2, h1, h2


In fact, the use tactic automatically tries to use available assumptions as well.

example : ∃ x : ℝ, 2 < x ∧ x < 3 := by
have h : 2 < (5 : ℝ) / 2 ∧ (5 : ℝ) / 2 < 3 := by norm_num
use 5 / 2


Alternatively, we can use Lean’s anonymous constructor notation to construct a proof of an existential quantifier.

example : ∃ x : ℝ, 2 < x ∧ x < 3 :=
have h : 2 < (5 : ℝ) / 2 ∧ (5 : ℝ) / 2 < 3 := by norm_num
⟨5 / 2, h⟩


Notice that there is no by; here we are giving an explicit proof term. The left and right angle brackets, which can be entered as \< and \> respectively, tell Lean to put together the given data using whatever construction is appropriate for the current goal. We can use the notation without going first into tactic mode:

example : ∃ x : ℝ, 2 < x ∧ x < 3 :=
⟨5 / 2, by norm_num⟩


So now we know how to prove an exists statement. But how do we use one? If we know that there exists an object with a certain property, we should be able to give a name to an arbitrary one and reason about it. For example, remember the predicates FnUb f a and FnLb f a from the last section, which say that a is an upper bound or lower bound on f, respectively. We can use the existential quantifier to say that “f is bounded” without specifying the bound:

def FnUb (f : ℝ → ℝ) (a : ℝ) : Prop :=
∀ x, f x ≤ a

def FnLb (f : ℝ → ℝ) (a : ℝ) : Prop :=
∀ x, a ≤ f x

def FnHasUb (f : ℝ → ℝ) :=
∃ a, FnUb f a

def FnHasLb (f : ℝ → ℝ) :=
∃ a, FnLb f a


We can use the theorem FnUb_add from the last section to prove that if f and g have upper bounds, then so does fun x ↦ f x + g x.

variable {f g : ℝ → ℝ}

example (ubf : FnHasUb f) (ubg : FnHasUb g) : FnHasUb fun x ↦ f x + g x := by
rcases ubf with ⟨a, ubfa⟩
rcases ubg with ⟨b, ubgb⟩
use a + b


The rcases tactic unpacks the information in the existential quantifier. The annotations like ⟨a, ubfa⟩, written with the same angle brackets as the anonymous constructors, are known as patterns, and they describe the information that we expect to find when we unpack the main argument. Given the hypothesis ubf that there is an upper bound for f, rcases ubf with ⟨a, ubfa⟩ adds a new variable a for an upper bound to the context, together with the hypothesis ubfa that it has the given property. The goal is left unchanged; what has changed is that we can now use the new object and the new hypothesis to prove the goal. This is a common method of reasoning in mathematics: we unpack objects whose existence is asserted or implied by some hypothesis, and then use it to establish the existence of something else.

Try using this method to establish the following. You might find it useful to turn some of the examples from the last section into named theorems, as we did with fn_ub_add, or you can insert the arguments directly into the proofs.

example (lbf : FnHasLb f) (lbg : FnHasLb g) : FnHasLb fun x ↦ f x + g x := by
sorry

example {c : ℝ} (ubf : FnHasUb f) (h : c ≥ 0) : FnHasUb fun x ↦ c * f x := by
sorry


The “r” in rcases stands for “recursive,” because it allows us to use arbitrarily complex patterns to unpack nested data. The rintro tactic is a combination of intro and rcases:

example : FnHasUb f → FnHasUb g → FnHasUb fun x ↦ f x + g x := by
rintro ⟨a, ubfa⟩ ⟨b, ubgb⟩
exact ⟨a + b, fnUb_add ubfa ubgb⟩


In fact, Lean also supports a pattern-matching fun in expressions and proof terms:

example : FnHasUb f → FnHasUb g → FnHasUb fun x ↦ f x + g x :=
fun ⟨a, ubfa⟩ ⟨b, ubgb⟩ ↦ ⟨a + b, fnUb_add ubfa ubgb⟩


The task of unpacking information in a hypothesis is so important that Lean and Mathlib provide a number of ways to do it. For example, the obtain tactic provides suggestive syntax:

example (ubf : FnHasUb f) (ubg : FnHasUb g) : FnHasUb fun x ↦ f x + g x := by
obtain ⟨a, ubfa⟩ := ubf
obtain ⟨b, ubgb⟩ := ubg
exact ⟨a + b, fnUb_add ubfa ubgb⟩


Think of the first obtain instruction as matching the “contents” of ubf with the given pattern and assigning the components to the named variables. rcases and obtain are said to destruct their arguments, though there is a small difference in that rcases clears ubf from the context when it is done, whereas it is still present after obtain.

Lean also supports syntax that is similar to that used in other functional programming languages:

example (ubf : FnHasUb f) (ubg : FnHasUb g) : FnHasUb fun x ↦ f x + g x := by
cases ubf
case intro a ubfa =>
cases ubg
case intro b ubgb =>
exact ⟨a + b, fnUb_add ubfa ubgb⟩

example (ubf : FnHasUb f) (ubg : FnHasUb g) : FnHasUb fun x ↦ f x + g x := by
cases ubf
next a ubfa =>
cases ubg
next b ubgb =>
exact ⟨a + b, fnUb_add ubfa ubgb⟩

example (ubf : FnHasUb f) (ubg : FnHasUb g) : FnHasUb fun x ↦ f x + g x := by
match ubf, ubg with
| ⟨a, ubfa⟩, ⟨b, ubgb⟩ =>
exact ⟨a + b, fnUb_add ubfa ubgb⟩

example (ubf : FnHasUb f) (ubg : FnHasUb g) : FnHasUb fun x ↦ f x + g x :=
match ubf, ubg with
| ⟨a, ubfa⟩, ⟨b, ubgb⟩ =>
⟨a + b, fnUb_add ubfa ubgb⟩


In the first example, if you put your cursor after cases ubf, you will see that the tactic produces a single goal, which Lean has tagged intro. (The particular name chosen comes from the internal name for the axiomatic primitive that builds a proof of an existential statement.) The case tactic then names the components. The second example is similar, except using next instead of case means that you can avoid mentioning intro. The word match in the last two examples highlights that what we are doing here is what computer scientists call “pattern matching.” Notice that the third proof begins by by, after which the tactic version of match expects a tactic proof on the right side of the arrow. The last example is a proof term: there are no tactics in sight.

For the rest of this book, we will stick to rcases, rintros, and obtain, as the preferred ways of using an existential quantifier. But it can’t hurt to see the alternative syntax, especially if there is a chance you will find yourself in the company of computer scientists.

To illustrate one way that rcases can be used, we prove an old mathematical chestnut: if two integers x and y can each be written as a sum of two squares, then so can their product, x * y. In fact, the statement is true for any commutative ring, not just the integers. In the next example, rcases unpacks two existential quantifiers at once. We then provide the magic values needed to express x * y as a sum of squares as a list to the use statement, and we use ring to verify that they work.

variable {α : Type*} [CommRing α]

def SumOfSquares (x : α) :=
∃ a b, x = a ^ 2 + b ^ 2

theorem sumOfSquares_mul {x y : α} (sosx : SumOfSquares x) (sosy : SumOfSquares y) :
SumOfSquares (x * y) := by
rcases sosx with ⟨a, b, xeq⟩
rcases sosy with ⟨c, d, yeq⟩
rw [xeq, yeq]
use a * c - b * d, a * d + b * c
ring


This proof doesn’t provide much insight, but here is one way to motivate it. A Gaussian integer is a number of the form $$a + bi$$ where $$a$$ and $$b$$ are integers and $$i = \sqrt{-1}$$. The norm of the Gaussian integer $$a + bi$$ is, by definition, $$a^2 + b^2$$. So the norm of a Gaussian integer is a sum of squares, and any sum of squares can be expressed in this way. The theorem above reflects the fact that norm of a product of Gaussian integers is the product of their norms: if $$x$$ is the norm of $$a + bi$$ and $$y$$ in the norm of $$c + di$$, then $$xy$$ is the norm of $$(a + bi) (c + di)$$. Our cryptic proof illustrates the fact that the proof that is easiest to formalize isn’t always the most perspicuous one. In Section 6.3, we will provide you with the means to define the Gaussian integers and use them to provide an alternative proof.

The pattern of unpacking an equation inside an existential quantifier and then using it to rewrite an expression in the goal comes up often, so much so that the rcases tactic provides an abbreviation: if you use the keyword rfl in place of a new identifier, rcases does the rewriting automatically (this trick doesn’t work with pattern-matching lambdas).

theorem sumOfSquares_mul' {x y : α} (sosx : SumOfSquares x) (sosy : SumOfSquares y) :
SumOfSquares (x * y) := by
rcases sosx with ⟨a, b, rfl⟩
rcases sosy with ⟨c, d, rfl⟩
use a * c - b * d, a * d + b * c
ring


As with the universal quantifier, you can find existential quantifiers hidden all over if you know how to spot them. For example, divisibility is implicitly an “exists” statement.

example (divab : a ∣ b) (divbc : b ∣ c) : a ∣ c := by
rcases divab with ⟨d, beq⟩
rcases divbc with ⟨e, ceq⟩
rw [ceq, beq]
use d * e; ring


And once again, this provides a nice setting for using rcases with rfl. Try it out in the proof above. It feels pretty good!

Then try proving the following:

example (divab : a ∣ b) (divac : a ∣ c) : a ∣ b + c := by
sorry


For another important example, a function $$f : \alpha \to \beta$$ is said to be surjective if for every $$y$$ in the codomain, $$\beta$$, there is an $$x$$ in the domain, $$\alpha$$, such that $$f(x) = y$$. Notice that this statement includes both a universal and an existential quantifier, which explains why the next example makes use of both intro and use.

example {c : ℝ} : Surjective fun x ↦ x + c := by
intro x
use x - c
dsimp; ring


Try this example yourself using the theorem mul_div_cancel'.:

example {c : ℝ} (h : c ≠ 0) : Surjective fun x ↦ c * x := by
sorry


At this point, it is worth mentioning that there is a tactic, field_simp, that will often clear denominators in a useful way. It can be used in conjunction with the ring tactic.

example (x y : ℝ) (h : x - y ≠ 0) : (x ^ 2 - y ^ 2) / (x - y) = x + y := by
field_simp [h]
ring


The next example uses a surjectivity hypothesis by applying it to a suitable value. Note that you can use rcases with any expression, not just a hypothesis.

example {f : ℝ → ℝ} (h : Surjective f) : ∃ x, f x ^ 2 = 4 := by
rcases h 2 with ⟨x, hx⟩
use x
rw [hx]
norm_num


See if you can use these methods to show that the composition of surjective functions is surjective.

variable {α : Type*} {β : Type*} {γ : Type*}
variable {g : β → γ} {f : α → β}

example (surjg : Surjective g) (surjf : Surjective f) : Surjective fun x ↦ g (f x) := by
sorry


## 3.3. Negation

The symbol ¬ is meant to express negation, so ¬ x < y says that x is not less than y, ¬ x = y (or, equivalently, x ≠ y) says that x is not equal to y, and ¬ ∃ z, x < z ∧ z < y says that there does not exist a z strictly between x and y. In Lean, the notation ¬ A abbreviates A → False, which you can think of as saying that A implies a contradiction. Practically speaking, this means that you already know something about how to work with negations: you can prove ¬ A by introducing a hypothesis h : A and proving False, and if you have h : ¬ A and h' : A, then applying h to h' yields False.

To illustrate, consider the irreflexivity principle lt_irrefl for a strict order, which says that we have ¬ a < a for every a. The asymmetry principle lt_asymm says that we have a < b → ¬ b < a. Let’s show that lt_asymm follows from lt_irrefl.

example (h : a < b) : ¬b < a := by
intro h'
have : a < a := lt_trans h h'
apply lt_irrefl a this


This example introduces a couple of new tricks. First, when you use have without providing a label, Lean uses the name this, providing a convenient way to refer back to it. Because the proof is so short, we provide an explicit proof term. But what you should really be paying attention to in this proof is the result of the intro tactic, which leaves a goal of False, and the fact that we eventually prove False by applying lt_irrefl to a proof of a < a.

Here is another example, which uses the predicate FnHasUb defined in the last section, which says that a function has an upper bound.

example (h : ∀ a, ∃ x, f x > a) : ¬FnHasUb f := by
intro fnub
rcases fnub with ⟨a, fnuba⟩
rcases h a with ⟨x, hx⟩
have : f x ≤ a := fnuba x
linarith


Remember that it is often convenient to use linarith when a goal follows from linear equations and inequalities that in the context.

See if you can prove these in a similar way:

example (h : ∀ a, ∃ x, f x < a) : ¬FnHasLb f :=
sorry

example : ¬FnHasUb fun x ↦ x :=
sorry


Mathlib offers a number of useful theorems for relating orders and negations:

#check (not_le_of_gt : a > b → ¬a ≤ b)
#check (not_lt_of_ge : a ≥ b → ¬a < b)
#check (lt_of_not_ge : ¬a ≥ b → a < b)
#check (le_of_not_gt : ¬a > b → a ≤ b)


Recall the predicate Monotone f, which says that f is nondecreasing. Use some of the theorems just enumerated to prove the following:

example (h : Monotone f) (h' : f a < f b) : a < b := by
sorry

example (h : a ≤ b) (h' : f b < f a) : ¬Monotone f := by
sorry


We can show that the first example in the last snippet cannot be proved if we replace < by ≤. Notice that we can prove the negation of a universally quantified statement by giving a counterexample. Complete the proof.

example : ¬∀ {f : ℝ → ℝ}, Monotone f → ∀ {a b}, f a ≤ f b → a ≤ b := by
intro h
let f := fun x : ℝ ↦ (0 : ℝ)
have monof : Monotone f := by sorry
have h' : f 1 ≤ f 0 := le_refl _
sorry


This example introduces the let tactic, which adds a local definition to the context. If you put the cursor after the let command, in the goal window you will see that the definition f : ℝ → ℝ := fun x ↦ 0 has been added to the context. Lean will unfold the definition of f when it has to. In particular, when we prove f 1 ≤ f 0 with le_refl, Lean reduces f 1 and f 0 to 0.

Use le_of_not_gt to prove the following:

example (x : ℝ) (h : ∀ ε > 0, x < ε) : x ≤ 0 := by
sorry


Implicit in many of the proofs we have just done is the fact that if P is any property, saying that there is nothing with property P is the same as saying that everything fails to have property P, and saying that not everything has property P is equivalent to saying that something fails to have property P. In other words, all four of the following implications are valid (but one of them cannot be proved with what we explained so far):

variable {α : Type*} (P : α → Prop) (Q : Prop)

example (h : ¬∃ x, P x) : ∀ x, ¬P x := by
sorry

example (h : ∀ x, ¬P x) : ¬∃ x, P x := by
sorry

example (h : ¬∀ x, P x) : ∃ x, ¬P x := by
sorry

example (h : ∃ x, ¬P x) : ¬∀ x, P x := by
sorry


The first, second, and fourth are straightforward to prove using the methods you have already seen. We encourage you to try it. The third is more difficult, however, because it concludes that an object exists from the fact that its nonexistence is contradictory. This is an instance of classical mathematical reasoning. We can use proof by contradiction to prove the third implication as follows.

example (h : ¬∀ x, P x) : ∃ x, ¬P x := by
by_contra h'
apply h
intro x
show P x
by_contra h''
exact h' ⟨x, h''⟩


Make sure you understand how this works. The by_contra tactic allows us to prove a goal Q by assuming ¬ Q and deriving a contradiction. In fact, it is equivalent to using the equivalence not_not : ¬ ¬ Q ↔ Q. Confirm that you can prove the forward direction of this equivalence using by_contra, while the reverse direction follows from the ordinary rules for negation.

example (h : ¬¬Q) : Q := by
sorry

example (h : Q) : ¬¬Q := by
sorry


Use proof by contradiction to establish the following, which is the converse of one of the implications we proved above. (Hint: use intro first.)

example (h : ¬FnHasUb f) : ∀ a, ∃ x, f x > a := by
sorry


It is often tedious to work with compound statements with a negation in front, and it is a common mathematical pattern to replace such statements with equivalent forms in which the negation has been pushed inward. To facilitate this, Mathlib offers a push_neg tactic, which restates the goal in this way. The command push_neg at h restates the hypothesis h.

example (h : ¬∀ a, ∃ x, f x > a) : FnHasUb f := by
push_neg at h
exact h

example (h : ¬FnHasUb f) : ∀ a, ∃ x, f x > a := by
dsimp only [FnHasUb, FnUb] at h
push_neg at h
exact h


In the second example, we use dsimp to expand the definitions of FnHasUb and FnUb. (We need to use dsimp rather than rw to expand FnUb, because it appears in the scope of a quantifier.) You can verify that in the examples above with ¬∃ x, P x and ¬∀ x, P x, the push_neg tactic does the expected thing. Without even knowing how to use the conjunction symbol, you should be able to use push_neg to prove the following:

example (h : ¬Monotone f) : ∃ x y, x ≤ y ∧ f y < f x := by
sorry


Mathlib also has a tactic, contrapose, which transforms a goal A → B to ¬B → ¬A. Similarly, given a goal of proving B from hypothesis h : A, contrapose h leaves you with a goal of proving ¬A from hypothesis ¬B. Using contrapose! instead of contrapose applies push_neg to the goal and the relevant hypothesis as well.

example (h : ¬FnHasUb f) : ∀ a, ∃ x, f x > a := by
contrapose! h
exact h

example (x : ℝ) (h : ∀ ε > 0, x ≤ ε) : x ≤ 0 := by
contrapose! h
use x / 2
constructor <;> linarith


We have not yet explained the constructor command or the use of the semicolon after it, but we will do that in the next section.

We close this section with the principle of ex falso, which says that anything follows from a contradiction. In Lean, this is represented by False.elim, which establishes False → P for any proposition P. This may seem like a strange principle, but it comes up fairly often. We often prove a theorem by splitting on cases, and sometimes we can show that one of the cases is contradictory. In that case, we need to assert that the contradiction establishes the goal so we can move on to the next one. (We will see instances of reasoning by cases in Section 3.5.)

Lean provides a number of ways of closing a goal once a contradiction has been reached.

example (h : 0 < 0) : a > 37 := by
exfalso
apply lt_irrefl 0 h

example (h : 0 < 0) : a > 37 :=
absurd h (lt_irrefl 0)

example (h : 0 < 0) : a > 37 := by
have h' : ¬0 < 0 := lt_irrefl 0


The exfalso tactic replaces the current goal with the goal of proving False. Given h : P and h' : ¬ P, the term absurd h h' establishes any proposition. Finally, the contradiction tactic tries to close a goal by finding a contradiction in the hypotheses, such as a pair of the form h : P and h' : ¬ P. Of course, in this example, linarith also works.

## 3.4. Conjunction and Iff

You have already seen that the conjunction symbol, ∧, is used to express “and.” The constructor tactic allows you to prove a statement of the form A ∧ B by proving A and then proving B.

example {x y : ℝ} (h₀ : x ≤ y) (h₁ : ¬y ≤ x) : x ≤ y ∧ x ≠ y := by
constructor
· assumption
intro h
apply h₁
rw [h]


In this example, the assumption tactic tells Lean to find an assumption that will solve the goal. Notice that the final rw finishes the goal by applying the reflexivity of ≤. The following are alternative ways of carrying out the previous examples using the anonymous constructor angle brackets. The first is a slick proof-term version of the previous proof, which drops into tactic mode at the keyword by.

example {x y : ℝ} (h₀ : x ≤ y) (h₁ : ¬y ≤ x) : x ≤ y ∧ x ≠ y :=
⟨h₀, fun h ↦ h₁ (by rw [h])⟩

example {x y : ℝ} (h₀ : x ≤ y) (h₁ : ¬y ≤ x) : x ≤ y ∧ x ≠ y :=
have h : x ≠ y := by
contrapose! h₁
rw [h₁]
⟨h₀, h⟩


Using a conjunction instead of proving one involves unpacking the proofs of the two parts. You can use the rcases tactic for that, as well as rintro or a pattern-matching fun, all in a manner similar to the way they are used with the existential quantifier.

example {x y : ℝ} (h : x ≤ y ∧ x ≠ y) : ¬y ≤ x := by
rcases h with ⟨h₀, h₁⟩
contrapose! h₁
exact le_antisymm h₀ h₁

example {x y : ℝ} : x ≤ y ∧ x ≠ y → ¬y ≤ x := by
rintro ⟨h₀, h₁⟩ h'
exact h₁ (le_antisymm h₀ h')

example {x y : ℝ} : x ≤ y ∧ x ≠ y → ¬y ≤ x :=
fun ⟨h₀, h₁⟩ h' ↦ h₁ (le_antisymm h₀ h')


In analogy to the obtain tactic, which we used with the existential quantifier, there is also a pattern-matching have:

example {x y : ℝ} (h : x ≤ y ∧ x ≠ y) : ¬y ≤ x := by
have ⟨h₀, h₁⟩ := h
contrapose! h₁
exact le_antisymm h₀ h₁


In contrast to rcases, here the have tactic leaves h in the context. And even though we won’t use them, once again we have the computer scientists’ pattern-matching syntax:

example {x y : ℝ} (h : x ≤ y ∧ x ≠ y) : ¬y ≤ x := by
cases h
case intro h₀ h₁ =>
contrapose! h₁
exact le_antisymm h₀ h₁

example {x y : ℝ} (h : x ≤ y ∧ x ≠ y) : ¬y ≤ x := by
cases h
next h₀ h₁ =>
contrapose! h₁
exact le_antisymm h₀ h₁

example {x y : ℝ} (h : x ≤ y ∧ x ≠ y) : ¬y ≤ x := by
match h with
| ⟨h₀, h₁⟩ =>
contrapose! h₁
exact le_antisymm h₀ h₁


In contrast to using an existential quantifier, you can also extract proofs of the two components of a hypothesis h : A ∧ B by writing h.left and h.right, or, equivalently, h.1 and h.2.

example {x y : ℝ} (h : x ≤ y ∧ x ≠ y) : ¬y ≤ x := by
intro h'
apply h.right
exact le_antisymm h.left h'

example {x y : ℝ} (h : x ≤ y ∧ x ≠ y) : ¬y ≤ x :=
fun h' ↦ h.right (le_antisymm h.left h')


Try using these techniques to come up with various ways of proving of the following:

example {m n : ℕ} (h : m ∣ n ∧ m ≠ n) : m ∣ n ∧ ¬n ∣ m :=
sorry


You can nest uses of ∃ and ∧ with anonymous constructors, rintro, and rcases.

example : ∃ x : ℝ, 2 < x ∧ x < 4 :=
⟨5 / 2, by norm_num, by norm_num⟩

example (x y : ℝ) : (∃ z : ℝ, x < z ∧ z < y) → x < y := by
rintro ⟨z, xltz, zlty⟩
exact lt_trans xltz zlty

example (x y : ℝ) : (∃ z : ℝ, x < z ∧ z < y) → x < y :=
fun ⟨z, xltz, zlty⟩ ↦ lt_trans xltz zlty


You can also use the use tactic:

example : ∃ x : ℝ, 2 < x ∧ x < 4 := by
use 5 / 2
constructor <;> norm_num

example : ∃ m n : ℕ, 4 < m ∧ m < n ∧ n < 10 ∧ Nat.Prime m ∧ Nat.Prime n := by
use 5
use 7
norm_num

example {x y : ℝ} : x ≤ y ∧ x ≠ y → x ≤ y ∧ ¬y ≤ x := by
rintro ⟨h₀, h₁⟩
use h₀
exact fun h' ↦ h₁ (le_antisymm h₀ h')


In the first example, the semicolon after the constructor command tells Lean to use the norm_num tactic on both of the goals that result.

In Lean, A ↔ B is not defined to be (A → B) ∧ (B → A), but it could have been, and it behaves roughly the same way. You have already seen that you can write h.mp and h.mpr or h.1 and h.2 for the two directions of h : A ↔ B. You can also use cases and friends. To prove an if-and-only-if statement, you can uses constructor or angle brackets, just as you would if you were proving a conjunction.

example {x y : ℝ} (h : x ≤ y) : ¬y ≤ x ↔ x ≠ y := by
constructor
· contrapose!
rintro rfl
rfl
contrapose!
exact le_antisymm h

example {x y : ℝ} (h : x ≤ y) : ¬y ≤ x ↔ x ≠ y :=
⟨fun h₀ h₁ ↦ h₀ (by rw [h₁]), fun h₀ h₁ ↦ h₀ (le_antisymm h h₁)⟩


The last proof term is inscrutable. Remember that you can use underscores while writing an expression like that to see what Lean expects.

Try out the various techniques and gadgets you have just seen in order to prove the following:

example {x y : ℝ} : x ≤ y ∧ ¬y ≤ x ↔ x ≤ y ∧ x ≠ y :=
sorry


For a more interesting exercise, show that for any two real numbers x and y, x^2 + y^2 = 0 if and only if x = 0 and y = 0. We suggest proving an auxiliary lemma using linarith, pow_two_nonneg, and pow_eq_zero.

theorem aux {x y : ℝ} (h : x ^ 2 + y ^ 2 = 0) : x = 0 :=
have h' : x ^ 2 = 0 := by sorry
pow_eq_zero h'

example (x y : ℝ) : x ^ 2 + y ^ 2 = 0 ↔ x = 0 ∧ y = 0 :=
sorry


In Lean, bi-implication leads a double-life. You can treat it like a conjunction and use its two parts separately. But Lean also knows that it is a reflexive, symmetric, and transitive relation between propositions, and you can also use it with calc and rw. It is often convenient to rewrite a statement to an equivalent one. In the next example, we use abs_lt to replace an expression of the form |x| < y by the equivalent expression - y < x ∧ x < y, and in the one after that we use Nat.dvd_gcd_iff to replace an expression of the form m ∣ Nat.gcd n k by the equivalent expression m ∣ n ∧ m ∣ k.

example (x : ℝ) : |x + 3| < 5 → -8 < x ∧ x < 2 := by
rw [abs_lt]
intro h
constructor <;> linarith

example : 3 ∣ Nat.gcd 6 15 := by
rw [Nat.dvd_gcd_iff]
constructor <;> norm_num


See if you can use rw with the theorem below to provide a short proof that negation is not a nondecreasing function. (Note that push_neg won’t unfold definitions for you, so the rw [Monotone] in the proof of the theorem is needed.)

theorem not_monotone_iff {f : ℝ → ℝ} : ¬Monotone f ↔ ∃ x y, x ≤ y ∧ f x > f y := by
rw [Monotone]
push_neg
rfl

example : ¬Monotone fun x : ℝ ↦ -x := by
sorry


The remaining exercises in this section are designed to give you some more practice with conjunction and bi-implication. Remember that a partial order is a binary relation that is transitive, reflexive, and antisymmetric. An even weaker notion sometimes arises: a preorder is just a reflexive, transitive relation. For any pre-order ≤, Lean axiomatizes the associated strict pre-order by a < b ↔ a ≤ b ∧ ¬ b ≤ a. Show that if ≤ is a partial order, then a < b is equivalent to a ≤ b ∧ a ≠ b:

variable {α : Type*} [PartialOrder α]
variable (a b : α)

example : a < b ↔ a ≤ b ∧ a ≠ b := by
rw [lt_iff_le_not_le]
sorry


Beyond logical operations, you do not need anything more than le_refl and le_trans. Show that even in the case where ≤ is only assumed to be a preorder, we can prove that the strict order is irreflexive and transitive. In the second example, for convenience, we use the simplifier rather than rw to express < in terms of ≤ and ¬. We will come back to the simplifier later, but here we are only relying on the fact that it will use the indicated lemma repeatedly, even if it needs to be instantiated to different values.

variable {α : Type*} [Preorder α]
variable (a b c : α)

example : ¬a < a := by
rw [lt_iff_le_not_le]
sorry

example : a < b → b < c → a < c := by
simp only [lt_iff_le_not_le]
sorry


## 3.5. Disjunction

The canonical way to prove a disjunction A ∨ B is to prove A or to prove B. The left tactic chooses A, and the right tactic chooses B.

variable {x y : ℝ}

example (h : y > x ^ 2) : y > 0 ∨ y < -1 := by
left
linarith [pow_two_nonneg x]

example (h : -y > x ^ 2 + 1) : y > 0 ∨ y < -1 := by
right
linarith [pow_two_nonneg x]


We cannot use an anonymous constructor to construct a proof of an “or” because Lean would have to guess which disjunct we are trying to prove. When we write proof terms we can use Or.inl and Or.inr instead to make the choice explicitly. Here, inl is short for “introduction left” and inr is short for “introduction right.”

example (h : y > 0) : y > 0 ∨ y < -1 :=
Or.inl h

example (h : y < -1) : y > 0 ∨ y < -1 :=
Or.inr h


It may seem strange to prove a disjunction by proving one side or the other. In practice, which case holds usually depends a case distinction that is implicit or explicit in the assumptions and the data. The rcases tactic allows us to make use of a hypothesis of the form A ∨ B. In contrast to the use of rcases with conjunction or an existential quantifier, here the rcases tactic produces two goals. Both have the same conclusion, but in the first case, A is assumed to be true, and in the second case, B is assumed to be true. In other words, as the name suggests, the rcases tactic carries out a proof by cases. As usual, we can tell Lean what names to use for the hypotheses. In the next example, we tell Lean to use the name h on each branch.

example : x < |y| → x < y ∨ x < -y := by
rcases le_or_gt 0 y with h | h
· rw [abs_of_nonneg h]
intro h; left; exact h
. rw [abs_of_neg h]
intro h; right; exact h


Notice that the pattern changes from ⟨h₀, h₁⟩ in the case of a conjunction to h₀ | h₁ in the case of a disjunction. Think of the first pattern as matching against data the contains both an h₀ and a h₁, whereas second pattern, with the bar, matches against data that contains either an h₀ or h₁. In this case, because the two goals are separate, we have chosen to use the same name, h, in each case.

The absolute value function is defined in such a way that we can immediately prove that x ≥ 0 implies |x| = x (this is the theorem abs_of_nonneg) and x < 0 implies |x| = -x (this is abs_of_neg). The expression le_or_gt 0 x establishes 0 ≤ x ∨ x < 0, allowing us to split on those two cases.

Lean also supports the computer scientists’ pattern-matching syntax for disjunction. Now the cases tactic is more attractive, because it allows us to name each case, and name the hypothesis that is introduced closer to where it is used.

example : x < |y| → x < y ∨ x < -y := by
cases le_or_gt 0 y
case inl h =>
rw [abs_of_nonneg h]
intro h; left; exact h
case inr h =>
rw [abs_of_neg h]
intro h; right; exact h


The names inl and inr are short for “intro left” and “intro right,” respectively. Using case has the advantage is that you can prove the cases in either order; Lean uses the tag to find the relevant goal. If you don’t care about that, you can use next, or match, or even a pattern-matching have.

example : x < |y| → x < y ∨ x < -y := by
cases le_or_gt 0 y
next h =>
rw [abs_of_nonneg h]
intro h; left; exact h
next h =>
rw [abs_of_neg h]
intro h; right; exact h

example : x < |y| → x < y ∨ x < -y := by
match le_or_gt 0 y with
| Or.inl h =>
rw [abs_of_nonneg h]
intro h; left; exact h
| Or.inr h =>
rw [abs_of_neg h]
intro h; right; exact h


In the case of the match, we need to use the full names Or.inl and Or.inr of the canonical ways to prove a disjunction. In this textbook, we will generally use rcases to split on the cases of a disjunction.

Try proving the triangle inequality using the two first two theorems in the next snippet. They are given the same names they have in Mathlib.

namespace MyAbs

theorem le_abs_self (x : ℝ) : x ≤ |x| := by
sorry

theorem neg_le_abs_self (x : ℝ) : -x ≤ |x| := by
sorry

theorem abs_add (x y : ℝ) : |x + y| ≤ |x| + |y| := by
sorry


In case you enjoyed these (pun intended) and you want more practice with disjunction, try these.

theorem lt_abs : x < |y| ↔ x < y ∨ x < -y := by
sorry

theorem abs_lt : |x| < y ↔ -y < x ∧ x < y := by
sorry


You can also use rcases and rintro with nested disjunctions. When these result in a genuine case split with multiple goals, the patterns for each new goal are separated by a vertical bar.

example {x : ℝ} (h : x ≠ 0) : x < 0 ∨ x > 0 := by
rcases lt_trichotomy x 0 with xlt | xeq | xgt
· left
exact xlt
. right; exact xgt


You can still nest patterns and use the rfl keyword to substitute equations:

example {m n k : ℕ} (h : m ∣ n ∨ m ∣ k) : m ∣ n * k := by
rcases h with ⟨a, rfl⟩ | ⟨b, rfl⟩
· rw [mul_assoc]
apply dvd_mul_right
. rw [mul_comm, mul_assoc]
apply dvd_mul_right


See if you can prove the following with a single (long) line. Use rcases to unpack the hypotheses and split on cases, and use a semicolon and linarith to solve each branch.

example {z : ℝ} (h : ∃ x y, z = x ^ 2 + y ^ 2 ∨ z = x ^ 2 + y ^ 2 + 1) : z ≥ 0 := by
sorry


On the real numbers, an equation x * y = 0 tells us that x = 0 or y = 0. In Mathlib, this fact is known as eq_zero_or_eq_zero_of_mul_eq_zero, and it is another nice example of how a disjunction can arise. See if you can use it to prove the following:

example {x : ℝ} (h : x ^ 2 = 1) : x = 1 ∨ x = -1 := by
sorry

example {x y : ℝ} (h : x ^ 2 = y ^ 2) : x = y ∨ x = -y := by
sorry


Remember that you can use the ring tactic to help with calculations.

In an arbitrary ring $$R$$, an element $$x$$ such that $$x y = 0$$ for some nonzero $$y$$ is called a left zero divisor, an element $$x$$ such that $$y x = 0$$ for some nonzero $$y$$ is called a right zero divisor, and an element that is either a left or right zero divisor is called simply a zero divisor. The theorem eq_zero_or_eq_zero_of_mul_eq_zero says that the real numbers have no nontrivial zero divisors. A commutative ring with this property is called an integral domain. Your proofs of the two theorems above should work equally well in any integral domain:

variable {R : Type*} [CommRing R] [IsDomain R]
variable (x y : R)

example (h : x ^ 2 = 1) : x = 1 ∨ x = -1 := by
sorry

example (h : x ^ 2 = y ^ 2) : x = y ∨ x = -y := by
sorry


In fact, if you are careful, you can prove the first theorem without using commutativity of multiplication. In that case, it suffices to assume that R is a Ring instead of an CommRing.

Sometimes in a proof we want to split on cases depending on whether some statement is true or not. For any proposition P, we can use em P : P ∨ ¬ P. The name em is short for “excluded middle.”

example (P : Prop) : ¬¬P → P := by
intro h
cases em P
· assumption


Alternatively, you can use the by_cases tactic.

example (P : Prop) : ¬¬P → P := by
intro h
by_cases h' : P
· assumption


Notice that the by_cases tactic lets you specify a label for the hypothesis that is introduced in each branch, in this case, h' : P in one and h' : ¬ P in the other. If you leave out the label, Lean uses h by default. Try proving the following equivalence, using by_cases to establish one direction.

example (P Q : Prop) : P → Q ↔ ¬P ∨ Q := by
sorry


## 3.6. Sequences and Convergence

We now have enough skills at our disposal to do some real mathematics. In Lean, we can represent a sequence $$s_0, s_1, s_2, \ldots$$ of real numbers as a function s : ℕ → ℝ. Such a sequence is said to converge to a number $$a$$ if for every $$\varepsilon > 0$$ there is a point beyond which the sequence remains within $$\varepsilon$$ of $$a$$, that is, there is a number $$N$$ such that for every $$n \ge N$$, $$| s_n - a | < \varepsilon$$. In Lean, we can render this as follows:

def ConvergesTo (s : ℕ → ℝ) (a : ℝ) :=
∀ ε > 0, ∃ N, ∀ n ≥ N, |s n - a| < ε


The notation ∀ ε > 0, ... is a convenient abbreviation for ∀ ε, ε > 0 → ..., and, similarly, ∀ n ≥ N, ... abbreviates ∀ n, n ≥ N →  .... And remember that ε > 0, in turn, is defined as 0 < ε, and n ≥ N is defined as N ≤ n.

In this section, we’ll establish some properties of convergence. But first, we will discuss three tactics for working with equality that will prove useful. The first, the ext tactic, gives us a way of proving that two functions are equal. Let $$f(x) = x + 1$$ and $$g(x) = 1 + x$$ be functions from reals to reals. Then, of course, $$f = g$$, because they return the same value for every $$x$$. The ext tactic enables us to prove an equation between functions by proving that their values are the same at all the values of their arguments.

example : (fun x y : ℝ ↦ (x + y) ^ 2) = fun x y : ℝ ↦ x ^ 2 + 2 * x * y + y ^ 2 := by
ext
ring


We’ll see later that ext is actually more general, and also one can specify the name of the variables that appear. For instance you can try to replace ext with ext u v in the above proof. The second tactic, the congr tactic, allows us to prove an equation between two expressions by reconciling the parts that are different:

example (a b : ℝ) : |a| = |a - b + b| := by
congr
ring


Here the congr tactic peels off the abs on each side, leaving us to prove a = a - b + b.

Finally, the convert tactic is used to apply a theorem to a goal when the conclusion of the theorem doesn’t quite match. For example, suppose we want to prove a < a * a from 1 < a. A theorem in the library, mul_lt_mul_right, will let us prove 1 * a < a * a. One possibility is to work backwards and rewrite the goal so that it has that form. Instead, the convert tactic lets us apply the theorem as it is, and leaves us with the task of proving the equations that are needed to make the goal match.

example {a : ℝ} (h : 1 < a) : a < a * a := by
convert (mul_lt_mul_right _).2 h
· rw [one_mul]
exact lt_trans zero_lt_one h


This example illustrates another useful trick: when we apply an expression with an underscore and Lean can’t fill it in for us automatically, it simply leaves it for us as another goal.

The following shows that any constant sequence $$a, a, a, \ldots$$ converges.

theorem convergesTo_const (a : ℝ) : ConvergesTo (fun x : ℕ ↦ a) a := by
intro ε εpos
use 0
intro n nge
rw [sub_self, abs_zero]
apply εpos


Lean has a tactic, simp, which can often save you the trouble of carrying out steps like rw [sub_self, abs_zero] by hand. We will tell you more about it soon.

For a more interesting theorem, let’s show that if s converges to a and t converges to b, then fun n ↦ s n + t n converges to a + b. It is helpful to have a clear pen-and-paper proof in mind before you start writing a formal one. Given ε greater than 0, the idea is to use the hypotheses to obtain an Ns such that beyond that point, s is within ε / 2 of a, and an Nt such that beyond that point, t is within ε / 2 of b. Then, whenever n is greater than or equal to the maximum of Ns and Nt, the sequence fun n ↦ s n + t n should be within ε of a + b. The following example begins to implement this strategy. See if you can finish it off.

theorem convergesTo_add {s t : ℕ → ℝ} {a b : ℝ}
(cs : ConvergesTo s a) (ct : ConvergesTo t b) :
ConvergesTo (fun n ↦ s n + t n) (a + b) := by
intro ε εpos
dsimp -- this line is not needed but cleans up the goal a bit.
have ε2pos : 0 < ε / 2 := by linarith
rcases cs (ε / 2) ε2pos with ⟨Ns, hs⟩
rcases ct (ε / 2) ε2pos with ⟨Nt, ht⟩
use max Ns Nt
sorry


As hints, you can use le_of_max_le_left and le_of_max_le_right, and norm_num can prove ε / 2 + ε / 2 = ε. Also, it is helpful to use the congr tactic to show that |s n + t n - (a + b)| is equal to |(s n - a) + (t n - b)|, since then you can use the triangle inequality. Notice that we marked all the variables s, t, a, and b implicit because they can be inferred from the hypotheses.

Proving the same theorem with multiplication in place of addition is tricky. We will get there by proving some auxiliary statements first. See if you can also finish off the next proof, which shows that if s converges to a, then fun n ↦ c * s n converges to c * a. It is helpful to split into cases depending on whether c is equal to zero or not. We have taken care of the zero case, and we have left you to prove the result with the extra assumption that c is nonzero.

theorem convergesTo_mul_const {s : ℕ → ℝ} {a : ℝ} (c : ℝ) (cs : ConvergesTo s a) :
ConvergesTo (fun n ↦ c * s n) (c * a) := by
by_cases h : c = 0
· convert convergesTo_const 0
· rw [h]
ring
rw [h]
ring
have acpos : 0 < |c| := abs_pos.mpr h
sorry


The next theorem is also independently interesting: it shows that a convergent sequence is eventually bounded in absolute value. We have started you off; see if you can finish it.

theorem exists_abs_le_of_convergesTo {s : ℕ → ℝ} {a : ℝ} (cs : ConvergesTo s a) :
∃ N b, ∀ n, N ≤ n → |s n| < b := by
rcases cs 1 zero_lt_one with ⟨N, h⟩
use N, |a| + 1
sorry


In fact, the theorem could be strengthened to assert that there is a bound b that holds for all values of n. But this version is strong enough for our purposes, and we will see at the end of this section that it holds more generally.

The next lemma is auxiliary: we prove that if s converges to a and t converges to 0, then fun n ↦ s n * t n converges to 0. To do so, we use the previous theorem to find a B that bounds s beyond some point N₀. See if you can understand the strategy we have outlined and finish the proof.

theorem aux {s t : ℕ → ℝ} {a : ℝ} (cs : ConvergesTo s a) (ct : ConvergesTo t 0) :
ConvergesTo (fun n ↦ s n * t n) 0 := by
intro ε εpos
dsimp
rcases exists_abs_le_of_convergesTo cs with ⟨N₀, B, h₀⟩
have Bpos : 0 < B := lt_of_le_of_lt (abs_nonneg _) (h₀ N₀ (le_refl _))
have pos₀ : ε / B > 0 := div_pos εpos Bpos
rcases ct _ pos₀ with ⟨N₁, h₁⟩
sorry


If you have made it this far, congratulations! We are now within striking distance of our theorem. The following proof finishes it off.

theorem convergesTo_mul {s t : ℕ → ℝ} {a b : ℝ}
(cs : ConvergesTo s a) (ct : ConvergesTo t b) :
ConvergesTo (fun n ↦ s n * t n) (a * b) := by
have h₁ : ConvergesTo (fun n ↦ s n * (t n + -b)) 0 := by
apply aux cs
ring
have := convergesTo_add h₁ (convergesTo_mul_const b cs)
convert convergesTo_add h₁ (convergesTo_mul_const b cs) using 1
· ext; ring
ring


For another challenging exercise, try filling out the following sketch of a proof that limits are unique. (If you are feeling bold, you can delete the proof sketch and try proving it from scratch.)

theorem convergesTo_unique {s : ℕ → ℝ} {a b : ℝ}
(sa : ConvergesTo s a) (sb : ConvergesTo s b) :
a = b := by
by_contra abne
have : |a - b| > 0 := by sorry
let ε := |a - b| / 2
have εpos : ε > 0 := by
change |a - b| / 2 > 0
linarith
rcases sa ε εpos with ⟨Na, hNa⟩
rcases sb ε εpos with ⟨Nb, hNb⟩
let N := max Na Nb
have absa : |s N - a| < ε := by sorry
have absb : |s N - b| < ε := by sorry
have : |a - b| < |a - b| := by sorry
exact lt_irrefl _ this


We close the section with the observation that our proofs can be generalized. For example, the only properties that we have used of the natural numbers is that their structure carries a partial order with min and max. You can check that everything still works if you replace ℕ everywhere by any linear order α:

variable {α : Type*} [LinearOrder α]

def ConvergesTo' (s : α → ℝ) (a : ℝ) :=
∀ ε > 0, ∃ N, ∀ n ≥ N, |s n - a| < ε


In Section 8.1, we will see that Mathlib has mechanisms for dealing with convergence in vastly more general terms, not only abstracting away particular features of the domain and codomain, but also abstracting over different types of convergence.