# 9. First Order Logic in Lean¶

## 9.1. Functions, Predicates, and Relations¶

In the last chapter, we discussed the language of first-order logic. We will see in the course of this book that Lean’s built-in logic is much more expressive; but it *includes* first-order logic, which is to say, anything that can be expressed (and proved) in first-order logic can be expressed (and proved) in Lean.

Lean is based on a foundational framework called *type theory*, in which every variable is assumed to range elements of some *type*. You can think of a type as being a “universe,” or a “domain of discourse,” in the sense of first-order logic.

For example, suppose we want to work with a first-order language with one constant symbol, one unary function symbol, one binary function symbol, one unary relation symbol, and one binary relation symbol. We can declare a new type `U`

(for “universe”) and the relevant symbols as follows:

```
constant U : Type
constant c : U
constant f : U → U
constant g : U → U → U
constant P : U → Prop
constant R : U → U → Prop
```

We can then use them as follows:

```
variables x y : U
#check c
#check f c
#check g x y
#check g x (f c)
#check P (g x (f c))
#check R x y
```

The `#check`

command tells us that the first four expressions have type `U`

, and that the last two have type `Prop`

. Roughly, this means that the first four expressions correspond to terms of first-order logic, and that the last two correspond to formulas.

Note all the following:

A unary function is represented as an object of type

`U → U`

and a binary function is represented as an object of type`U → U → U`

, using the same notation as for implication between propositions.We write, for example,

`f x`

to denote the result of applying`f`

to`x`

, and`g x y`

to denote the result of applying`g`

to`x`

and`y`

, again just as we did when using modus ponens for first-order logic. Parentheses are needed in the expression`g x (f c)`

to ensure that`f c`

is parsed as a single argument.A unary predicate is presented as an object of type

`U → Prop`

and a binary predicate is represented as an object of type`U → U → Prop`

. You can think of a binary relation`R`

as being a function that assumes two arguments in the universe,`U`

, and returns a proposition.We write

`P x`

to denote the assertion that`P`

holds of`x`

, and`R x y`

to denote that`R`

holds of`x`

and`y`

.

You may reasonably wonder what difference there is between a constant and a variable in Lean. The following declarations also work:

```
variable U : Type
variable c : U
variable f : U → U
variable g : U → U → U
variable P : U → Prop
variable R : U → U → Prop
variables x y : U
#check c
#check f c
#check g x y
#check g x (f c)
#check P (g x (f c))
#check R x y
```

Although the examples function in much the same way, the `constant`

and `variable`

commands do very different things. The `constant`

command declares a new object, axiomatically, and adds it to the list of objects Lean knows about. In contrast, when it is first executed, the `variable`

command does not create anything. Rather, it tells Lean that whenever we enter an expression using the corresponding identifier, it should create a temporary variable of the corresponding type.

Many types are already declared in Lean’s standard library. For example, there is a type written `nat`

or `ℕ`

, that denotes the natural numbers:

```
#check nat
#check ℕ
```

You can enter the unicode `ℕ`

with `\nat`

or `\N`

. The two expressions mean the same thing.

Using this built-in type, we can model the language of arithmetic, as described in the last chapter, as follows:

```
namespace hidden
constant mul : ℕ → ℕ → ℕ
constant add : ℕ → ℕ → ℕ
constant square : ℕ → ℕ
constant even : ℕ → Prop
constant odd : ℕ → Prop
constant prime : ℕ → Prop
constant divides : ℕ → ℕ → Prop
constant lt : ℕ → ℕ → Prop
constant zero : ℕ
constant one : ℕ
end hidden
```

We have used the `namespace`

command to avoid conflicts with identifiers that are already declared in the Lean library. (Outside the namespace, the constant `mul`

we just declared is named `hidden.mul`

.) We can again use the `#check`

command to try them out:

```
namespace hidden
constant mul : ℕ → ℕ → ℕ
constant add : ℕ → ℕ → ℕ
constant square : ℕ → ℕ
constant even : ℕ → Prop
constant odd : ℕ → Prop
constant prime : ℕ → Prop
constant divides : ℕ → ℕ → Prop
constant lt : ℕ → ℕ → Prop
constant zero : ℕ
constant one : ℕ
variables w x y z : ℕ
#check mul x y
#check add x y
#check square x
#check even x
end hidden
```

We can even declare infix notation of binary operations and relations:

```
infix + := add
infix * := mul
infix < := lt
```

(Getting notation for numerals `1`

, `2`

, `3`

, … is trickier.) With all this in place, the examples above can be rendered as follows:

```
#check even (x + y + z) ∧ prime ((x + one) * y * y)
#check ¬ (square (x + y * z) = w) ∨ x + y < z
#check x < y ∧ even x ∧ even y → x + one < y
```

In fact, all of the functions, predicates, and relations discussed here, except for the “square” function and “prime,” are defined in the core Lean library. They become available to us when we put the commands `import data.nat`

and `open nat`

at the top of a file in Lean.

```
import data.nat
open nat
constant square : ℕ → ℕ
constant prime : ℕ → Prop
constant even : ℕ → Prop
variables w x y z : ℕ
#check even (x + y + z) ∧ prime ((x + 1) * y * y)
#check ¬ (square (x + y * z) = w) ∨ x + y < z
#check x < y ∧ even x ∧ even y → x + 1 < y
```

Here, we declare the constants `square`

and `prime`

axiomatically, but refer to the other operations and predicates in the Lean library. In this book, we will often proceed in this way, telling you explicitly what facts from the library you should use for exercises.

Again, note the following aspects of syntax:

In contrast to ordinary mathematical notation, in Lean, functions are applied without parentheses or commas. For example, we write

`square x`

and`add x y`

instead of \(\mathit{square}(x)\) and \(\mathit{add}(x, y)\).The same holds for predicates and relations: we write

`even x`

and`lt x y`

instead of \(\mathit{even}(x)\) and \(\mathit{lt}(x, y)\), as one might do in symbolic logic.The notation

`add : ℕ → ℕ → ℕ`

indicates that addition assumes two arguments, both natural numbers, and returns a natural number.Similarly, the notation

`divides : ℕ → ℕ → Prop`

indicates that`divides`

is a binary relation, which assumes two natural numbers as arguments and forms a proposition. In other words,`divides x y`

expresses the assertion that`x`

divides`y`

.

Lean can help us distinguish between terms and formulas. If we `#check`

the expression `x + y + 1`

in Lean, we are told it has type `ℕ`

, which is to say, it denotes a natural number. If we `#check`

the expression `even (x + y + 1)`

, we are told that it has type `Prop`

, which is to say, it expresses a proposition.

In Chapter 7 we considered many-sorted logic, where one can have multiple universes. For example, we might want to use first-order logic for geometry, with quantifiers ranging over points and lines. In Lean, we can model this as by introducing a new type for each sort:

```
variables Point Line : Type
variable lies_on : Point → Line → Prop
```

We can then express that two distinct points determine a line as follows:

```
#check ∀ (p q : Point) (L M : Line),
p ≠ q → lies_on p L → lies_on q L → lies_on p M →
lies_on q M → L = M
```

Notice that we have followed the convention of using iterated implication rather than conjunction in the antecedent. In fact, Lean is smart enough to infer what sorts of objects `p`

, `q`

, `L`

, and `M`

are from the fact that they are used with the relation `lies_on`

, so we could have written, more simply, this:

```
#check ∀ p q L M, p ≠ q → lies_on p L → lies_on q L →
lies_on p M → lies_on q M → L = M
```

## 9.2. Using the Universal Quantifier¶

In Lean, you can enter the universal quantifier by writing `\all`

. The motivating examples from Section 7.1 are rendered as follows:

```
import data.nat
open nat
constant prime : ℕ → Prop
constant even : ℕ → Prop
constant odd : ℕ → Prop
#check ∀ x, (even x ∨ odd x) ∧ ¬ (even x ∧ odd x)
#check ∀ x, even x ↔ 2 ∣ x
#check ∀ x, even x → even (x^2)
#check ∀ x, even x ↔ odd (x + 1)
#check ∀ x, prime x ∧ x > 2 → odd x
#check ∀ x y z, x ∣ y → y ∣ z → x ∣ z
```

Remember that Lean expects a comma after the universal quantifier, and gives it the *widest* scope possible. For example, `∀ x, P ∨ Q`

is interpreted as `∀ x, (P ∨ Q)`

, and we would write `(∀ x, P) ∨ Q`

to limit the scope. If you prefer, you can use the plain ascii expression `forall`

instead of the unicode `∀`

.

In Lean, then, the pattern for proving a universal statement is rendered as follows:

```
variable U : Type
variable P : U → Prop
example : ∀ x, P x :=
assume x,
show P x, from sorry
```

Read `assume x`

as “fix an arbitrary value `x`

of `U`

.” Since we are allowed to rename bound variables at will, we can equivalently write either of the following:

```
variable U : Type
variable P : U → Prop
example : ∀ y, P y :=
assume x,
show P x, from sorry
example : ∀ x, P x :=
assume y,
show P y, from sorry
```

This constitutes the introduction rule for the universal quantifier. It is very similar to the introduction rule for implication: instead of using `assume`

to temporarily introduce an assumption, we use `assume`

to temporarily introduce a new object, `y`

. (In fact, `assume`

and `assume`

are both alternate syntax for a single internal construct in Lean, which can also be denoted by `λ`

.)

The elimination rule is, similarly, implemented as follows:

```
variable U : Type
variable P : U → Prop
variable h : ∀ x, P x
variable a : U
example : P a :=
show P a, from h a
```

Observe the notation: `P a`

is obtained by “applying” the hypothesis `h`

to `a`

. Once again, note the similarity to the elimination rule for implication.

Here is an example of how it is used:

```
variable U : Type
variables A B : U → Prop
example (h1 : ∀ x, A x → B x) (h2 : ∀ x, A x) : ∀ x, B x :=
assume y,
have h3 : A y, from h2 y,
have h4 : A y → B y, from h1 y,
show B y, from h4 h3
```

Here is an even shorter version of the same proof, where we avoid using `have`

:

```
example (h1 : ∀ x, A x → B x) (h2 : ∀ x, A x) : ∀ x, B x :=
assume y,
show B y, from h1 y (h2 y)
```

You should talk through the steps, here. Applying `h1`

to `y`

yields a proof of `A y → B y`

, which we then apply to `h2 y`

, which is a proof of `A y`

. The result is the proof of `B y`

that we are after.

In the last chapter, we considered the following proof in natural deduction:

Here is the same proof rendered in Lean:

```
variable U : Type
variables A B : U → Prop
example : (∀ x, A x) → (∀ x, B x) → (∀ x, A x ∧ B x) :=
assume hA : ∀ x, A x,
assume hB : ∀ x, B x,
assume y,
have Ay : A y, from hA y,
have By : B y, from hB y,
show A y ∧ B y, from and.intro Ay By
```

Here is an alternative version, using the “anonymous” versions of `have`

:

```
variable U : Type
variables A B : U → Prop
example : (∀ x, A x) → (∀ x, B x) → (∀ x, A x ∧ B x) :=
assume hA : ∀ x, A x,
assume hB : ∀ x, B x,
assume y,
have A y, from hA y,
have B y, from hB y,
show A y ∧ B y, from and.intro ‹A y› ‹B y›
```

The exercises below ask you to prove the barber paradox, which was discussed in the last chapter. You can do that using only propositional reasoning and the rules for the universal quantifier that we have just discussed.

## 9.3. Using the Existential Quantifier¶

In Lean, you can type the existential quantifier, `∃`

, by writing `\ex`

. If you prefer you can use the ascii equivalent, `exists`

. The introduction rule is `exists.intro`

and requires two arguments: a term, and a proof that that term satisfies the required property.

```
variable U : Type
variable P : U → Prop
example (y : U) (h : P y) : ∃ x, P x :=
exists.intro y h
```

The elimination rule for the existential quantifier is given by `exists.elim`

. It follows the form of the natural deduction rule: if we know `∃x, P x`

and we are trying to prove `Q`

, it suffices to introduce a new variable, `y`

, and prove `Q`

under the assumption that `P y`

holds.

```
variable U : Type
variable P : U → Prop
variable Q : Prop
example (h1 : ∃ x, P x) (h2 : ∀ x, P x → Q) : Q :=
exists.elim h1
(assume (y : U) (h : P y),
have h3 : P y → Q, from h2 y,
show Q, from h3 h)
```

As usual, we can leave off the information as to the data type of `y`

and the hypothesis `h`

after the `assume`

, since Lean can figure them out from the context. Deleting the `show`

and replacing `h3`

by its proof, `h2 y`

, yields a short (though virtually unreadable) proof of the conclusion.

```
example (h1 : ∃ x, P x) (h2 : ∀ x, P x → Q) : Q :=
exists.elim h1 (assume y h, h2 y h)
```

The following example uses both the introduction and the elimination rules for the existential quantifier.

```
variable U : Type
variables A B : U → Prop
example : (∃ x, A x ∧ B x) → ∃ x, A x :=
assume h1 : ∃ x, A x ∧ B x,
exists.elim h1
(assume y (h2 : A y ∧ B y),
have h3 : A y, from and.left h2,
show ∃ x, A x, from exists.intro y h3)
```

Notice the parentheses in the hypothesis; if we left them out, everything after the first `∃ x`

would be included in the scope of that quantifier. From the hypothesis, we obtain a `y`

that satisfies `A y ∧ B y`

, and hence `A y`

in particular. So `y`

is enough to witness the conclusion.

It is sometimes annoying to enclose the proof after an `exists.elim`

in parenthesis, as we did here with the `assume ... show`

block. To avoid that, we can use a bit of syntax from the programming world, and use a dollar sign instead. In Lean, an expression `f $ t`

means the same thing as `f (t)`

, with the advantage that we do not have to remember to close the parenthesis. With this gadget, we can write the proof above as follows:

```
variable U : Type
variables A B : U → Prop
example : (∃ x, A x ∧ B x) → ∃ x, A x :=
assume h1 : ∃ x, A x ∧ B x,
exists.elim h1 $
assume y (h2 : A y ∧ B y),
have h3 : A y, from and.left h2,
show ∃ x, A x, from exists.intro y h3
```

The following example is more involved:

```
example : (∃ x, A x ∨ B x) → (∃ x, A x) ∨ (∃ x, B x) :=
assume h1 : ∃ x, A x ∨ B x,
exists.elim h1 $
assume y (h2 : A y ∨ B y),
or.elim h2
(assume h3 : A y,
have h4 : ∃ x, A x, from exists.intro y h3,
show (∃ x, A x) ∨ (∃ x, B x), from or.inl h4)
(assume h3 : B y,
have h4 : ∃ x, B x, from exists.intro y h3,
show (∃ x, A x) ∨ (∃ x, B x), from or.inr h4)
```

Note again the placement of parentheses in the statement.

In the last chapter, we considered the following natural deduction proof:

Here is a proof of the same implication in Lean:

```
variable U : Type
variables A B : U → Prop
example : (∀ x, A x → ¬ B x) → ¬ ∃ x, A x ∧ B x :=
assume h1 : ∀ x, A x → ¬ B x,
assume h2 : ∃ x, A x ∧ B x,
exists.elim h2 $
assume x (h3 : A x ∧ B x),
have h4 : A x, from and.left h3,
have h5 : B x, from and.right h3,
have h6 : ¬ B x, from h1 x h4,
show false, from h6 h5
```

Here, we use `exists.elim`

to introduce a value `x`

satisfying `A x ∧ B x`

. The name is arbitrary; we could just as well have used `z`

:

```
example : (∀ x, A x → ¬ B x) → ¬ ∃ x, A x ∧ B x :=
assume h1 : ∀ x, A x → ¬ B x,
assume h2 : ∃ x, A x ∧ B x,
exists.elim h2 $
assume z (h3 : A z ∧ B z),
have h4 : A z, from and.left h3,
have h5 : B z, from and.right h3,
have h6 : ¬ B z, from h1 z h4,
show false, from h6 h5
```

Here is another example of the exists-elimination rule:

```
variable U : Type
variable u : U
variable P : Prop
example : (∃x : U, P) ↔ P :=
iff.intro
(assume h1 : ∃x, P,
exists.elim h1 $
assume x (h2 : P),
h2)
(assume h1 : P,
exists.intro u h1)
```

It is subtle: the proof does not go through if we do not declare a variable `u`

of type `U`

, even though `u`

does not appear in the statement of the theorem. The semantics of first-order logic, discussed in the next chapter, presuppose that the universe is nonempty. In Lean, however, it is possible for a type to be empty, and so the proof above depends on the fact that there is an element `u`

in `U`

.

These features are all illustrated in the following example:

```
variable U : Type
variables P R : U → Prop
variable Q : Prop
example (h1 : ∃x, P x ∧ R x) (h2 : ∀x, P x → R x → Q) : Q :=
let ⟨y, hPy, hRy⟩ := h1 in
show Q, from h2 y hPy hRy
```

## 9.4. Equality and calculational proofs¶

In Lean, reflexivity, symmetry, and transitivity are called `eq.refl`

, `eq.symm`

, and `eq.trans`

, and the second substitution rule is called `eq.subst`

. Their uses are illustrated below.

```
variable A : Type
variables x y z : A
variable P : A → Prop
example : x = x :=
show x = x, from eq.refl x
example : y = x :=
have h : x = y, from sorry,
show y = x, from eq.symm h
example : x = z :=
have h1 : x = y, from sorry,
have h2 : y = z, from sorry,
show x = z, from eq.trans h1 h2
example : P y :=
have h1 : x = y, from sorry,
have h2 : P x, from sorry,
show P y, from eq.subst h1 h2
```

The rule `eq.refl`

above assumes `x`

as an argument, because there is no hypothesis to infer it from. All the other rules assume their premises as arguments. Here is an example of equational reasoning:

```
variables (A : Type) (x y z : A)
example : y = x → y = z → x = z :=
assume h1 : y = x,
assume h2 : y = z,
have h3 : x = y, from eq.symm h1,
show x = z, from eq.trans h3 h2
```

This proof can be written more concisely:

```
example : y = x → y = z → x = z :=
assume h1 h2, eq.trans (eq.symm h1) h2
```

Because calculations are so important in mathematics, however, Lean provides more efficient ways of carrying them out. One method is to use the `rewrite`

tactic. Typing `begin`

and `end`

anywhere a proof is expected puts Lean into *tactic mode*, which provides an alternative way of writing a proof: rather than writing it directly, you provide Lean with a list of instructions that show Lean how to construct a proof of the statement in question. The statement to be proved is called the *goal*, and many instructions make progress by transforming the goal into something that is easier to prove. The `rewrite`

command, which carries out a substitution on the goal, is a good example. The previous example can be proved as follows:

```
example : y = x → y = z → x = z :=
assume h1 : y = x,
assume h2 : y = z,
show x = z,
begin
rewrite ←h1,
apply h2
end
```

If you put the cursor after the word `begin`

, Lean will tell you that the goal at that point is to prove `x = z`

. The first command changes the goal `x = z`

to `y = z`

; the left-facing arrow before `h1`

(which you can enter as `\<-`

) tells Lean to use the equation in the reverse direction. If you put the cursor after the comma, Lean shows you the new goal, `y = z`

. The `apply`

command uses `h2`

to complete the proof.

An alternative is to rewrite the goal using `h1`

and `h2`

, which reduces the goal to `x = x`

. When that happens, `rewrite`

automatically applies reflexivity.

```
example : y = x → y = z → x = z :=
assume h1 : y = x,
assume h2 : y = z,
show x = z,
begin
rw ←h1,
rw h2
end
```

In fact, a sequence of rewrites can be combined, using square brackets:

```
example : y = x → y = z → x = z :=
assume h1 : y = x,
assume h2 : y = z,
show x = z,
begin
rw [←h1, h2]
end
```

And when you reduce a proof to a single tactic, you can use `by`

instead of `begin ... end`

.

```
example : y = x → y = z → x = z :=
assume h1 : y = x,
assume h2 : y = z,
show x = z, by rw [←h1, h2]
```

If you put the cursor after the `←h1`

, Lean shows you the goal at that point.

We will see in the coming chapters that in ordinary mathematical proofs, one commonly carries out calculations in a format like this:

Lean has a mechanism to model such calculational proofs. Whenever a proof of an equation is expected, you can provide a proof using the identifier `calc`

, following by a chain of equalities and justification, in the following form:

```
calc
e1 = e2 : justification 1
... = e3 : justification 2
... = e4 : justification 3
... = e5 : justification 4
```

The chain can go on as long as needed, and in this example the result is a proof of `e1 = e5`

. Each justification is the name of the assumption or theorem that is used. For example, the previous proof could be written as follows:

```
example : y = x → y = z → x = z :=
assume h1 : y = x,
assume h2 : y = z,
calc
x = y : eq.symm h1
... = z : h2
```

As usual, the syntax is finicky; notice that there are no commas in the `calc`

expression, and the colons and dots need to be entered exactly in that form. All that varies are the expressions `e1, e2, e3, ...`

and the justifications themselves.

The `calc`

environment is most powerful when used in conjunction with `rewrite`

, since we can then rewrite expressions with facts from the library. For example, Lean’s library has a number of basic identities for the integers, such as these:

```
variables x y z : int
example : x + 0 = x :=
add_zero x
example : 0 + x = x :=
zero_add x
example : (x + y) + z = x + (y + z) :=
add_assoc x y z
example : x + y = y + x :=
add_comm x y
example : (x * y) * z = x * (y * z) :=
mul_assoc x y z
example : x * y = y * x :=
mul_comm x y
example : x * (y + z) = x * y + x * z :=
left_distrib x y z
example : (x + y) * z = x * z + y * z :=
right_distrib x y z
```

You can also write the type of integers as `ℤ`

, entered with either `\Z`

or `\int`

. Notice that, for example, `add_comm`

is the theorem `∀ x y, x + y = y + x`

. So to instantiate it to `s + t = t + s`

, you write `add_comm s t`

. Using these axioms, here is the calculation above rendered in Lean, as a theorem about the integers:

```
example (x y z : int) : (x + y) + z = (x + z) + y :=
calc
(x + y) + z = x + (y + z) : add_assoc x y z
... = x + (z + y) : eq.subst (add_comm y z) rfl
... = (x + z) + y : eq.symm (add_assoc x z y)
```

Using `rewrite`

is more efficient, though at times we have to provide information to specify where the rules are used:

```
example (x y z : int) : (x + y) + z = (x + z) + y :=
calc
(x + y) + z = x + (y + z) : by rw add_assoc
... = x + (z + y) : by rw [add_comm y z]
... = (x + z) + y : by rw add_assoc
```

In that case, we can use a single `rewrite`

:

```
example (x y z : int) : (x + y) + z = (x + z) + y :=
by rw [add_assoc, add_comm y z, add_assoc]
```

If you #check the proof before the sequence of `rewrites`

is sufficient, the error message will display the remaining goal.

Here is another example:

```
variables a b d c : int
example : (a + b) * (c + d) = a * c + b * c + a * d + b * d :=
calc
(a + b) * (c + d) = (a + b) * c + (a + b) * d : by rw left_distrib
... = (a * c + b * c) + (a + b) * d : by rw right_distrib
... = (a * c + b * c) + (a * d + b * d) : by rw right_distrib
... = a * c + b * c + a * d + b * d : by rw ←add_assoc
```

Once again, we can get by with a shorter proof:

```
example : (a + b) * (c + d) = a * c + b * c + a * d + b * d :=
by rw [left_distrib, right_distrib, right_distrib, ←add_assoc]
```

## 9.5. Exercises¶

Fill in the

`sorry`

.section variable A : Type variable f : A → A variable P : A → Prop variable h : ∀ x, P x → P (f x) -- Show the following: example : ∀ y, P y → P (f (f y)) := sorry end

Fill in the

`sorry`

.section variable U : Type variables A B : U → Prop example : (∀ x, A x ∧ B x) → ∀ x, A x := sorry end

Fill in the

`sorry`

.section variable U : Type variables A B C : U → Prop variable h1 : ∀ x, A x ∨ B x variable h2 : ∀ x, A x → C x variable h3 : ∀ x, B x → C x example : ∀ x, C x := sorry end

Fill in the

`sorry`

’s below, to prove the barber paradox.open classical -- not needed, but you can use it -- This is an exercise from Chapter 4. Use it as an axiom here. axiom not_iff_not_self (P : Prop) : ¬ (P ↔ ¬ P) example (Q : Prop) : ¬ (Q ↔ ¬ Q) := not_iff_not_self Q section variable Person : Type variable shaves : Person → Person → Prop variable barber : Person variable h : ∀ x, shaves barber x ↔ ¬ shaves x x -- Show the following: example : false := sorry end

Fill in the

`sorry`

.section variable U : Type variables A B : U → Prop example : (∃ x, A x) → ∃ x, A x ∨ B x := sorry end

Fill in the

`sorry`

.section variable U : Type variables A B : U → Prop variable h1 : ∀ x, A x → B x variable h2 : ∃ x, A x example : ∃ x, B x := sorry end

Fill in the

`sorry`

.variable U : Type variables A B C : U → Prop example (h1 : ∃ x, A x ∧ B x) (h2 : ∀ x, B x → C x) : ∃ x, A x ∧ C x := sorry

Complete these proofs.

variable U : Type variables A B C : U → Prop example : (¬ ∃ x, A x) → ∀ x, ¬ A x := sorry example : (∀ x, ¬ A x) → ¬ ∃ x, A x := sorry

Fill in the

`sorry`

.variable U : Type variables R : U → U → Prop example : (∃ x, ∀ y, R x y) → ∀ y, ∃ x, R x y := sorry

The following exercise shows that in the presence of reflexivity, the rules for symmetry and transitivity are equivalent to a single rule.

theorem foo {A : Type} {a b c : A} : a = b → c = b → a = c := sorry -- notice that you can now use foo as a rule. The curly braces mean that -- you do not have to give A, a, b, or c section variable A : Type variables a b c : A example (h1 : a = b) (h2 : c = b) : a = c := foo h1 h2 end section variable {A : Type} variables {a b c : A} -- replace the sorry with a proof, using foo and rfl, without using eq.symm. theorem my_symm (h : b = a) : a = b := sorry -- now use foo and my_symm to prove transitivity theorem my_trans (h1 : a = b) (h2 : b = c) : a = c := sorry end

Replace each

`sorry`

below by the correct axiom from the list.-- these are the axioms for a commutative ring #check @add_assoc #check @add_comm #check @add_zero #check @zero_add #check @mul_assoc #check @mul_comm #check @mul_one #check @one_mul #check @left_distrib #check @right_distrib #check @add_left_neg #check @add_right_neg #check @sub_eq_add_neg variables x y z : int theorem t1 : x - x = 0 := calc x - x = x + -x : by rw sub_eq_add_neg ... = 0 : by rw add_right_neg theorem t2 (h : x + y = x + z) : y = z := calc y = 0 + y : by rw zero_add ... = (-x + x) + y : by rw add_left_neg ... = -x + (x + y) : by rw add_assoc ... = -x + (x + z) : by rw h ... = (-x + x) + z : by rw add_assoc ... = 0 + z : by rw add_left_neg ... = z : by rw zero_add theorem t3 (h : x + y = z + y) : x = z := calc x = x + 0 : sorry ... = x + (y + -y) : sorry ... = (x + y) + -y : sorry ... = (z + y) + -y : by rw h ... = z + (y + -y) : sorry ... = z + 0 : sorry ... = z : sorry theorem t4 (h : x + y = 0) : x = -y := calc x = x + 0 : by rw add_zero ... = x + (y + -y) : by rw add_right_neg ... = (x + y) + -y : by rw add_assoc ... = 0 + -y : by rw h ... = -y : by rw zero_add theorem t5 : x * 0 = 0 := have h1 : x * 0 + x * 0 = x * 0 + 0, from calc x * 0 + x * 0 = x * (0 + 0) : sorry ... = x * 0 : sorry ... = x * 0 + 0 : sorry, show x * 0 = 0, from t2 _ _ _ h1 theorem t6 : x * (-y) = -(x * y) := have h1 : x * (-y) + x * y = 0, from calc x * (-y) + x * y = x * (-y + y) : sorry ... = x * 0 : sorry ... = 0 : by rw t5 x, show x * (-y) = -(x * y), from t4 _ _ h1 theorem t7 : x + x = 2 * x := calc x + x = 1 * x + 1 * x : by rw one_mul ... = (1 + 1) * x : sorry ... = 2 * x : rfl