l3lab/ntp-mathlib huggingface dataset:
theorem ne_zero_of_coprime (hc : Int.gcd x y = 1) : z ≠ 0 := by
  suffices 0 < z * z by
    rintro rfl
    norm_num at this
  rw [← h.eq, ← sq, ← sq]
  

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
⊢ 0 < x ^ 2 + y ^ 2

Next tactic:
have hc' : Int.gcd x y ≠ 0 := by
    rw [hc]
    exact one_ne_zero
theorem ne_zero_of_coprime (hc : Int.gcd x y = 1) : z ≠ 0 := by
  suffices 0 < z * z by
    rintro rfl
    norm_num at this
  rw [← h.eq, ← sq, ← sq]
  have hc' : Int.gcd x y ≠ 0 := by
    

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
⊢ Int.gcd x y ≠ 0

Next tactic:
rw [hc]
theorem ne_zero_of_coprime (hc : Int.gcd x y = 1) : z ≠ 0 := by
  suffices 0 < z * z by
    rintro rfl
    norm_num at this
  rw [← h.eq, ← sq, ← sq]
  have hc' : Int.gcd x y ≠ 0 := by
    rw [hc]
    

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
⊢ 1 ≠ 0

Next tactic:
exact one_ne_zero
theorem ne_zero_of_coprime (hc : Int.gcd x y = 1) : z ≠ 0 := by
  suffices 0 < z * z by
    rintro rfl
    norm_num at this
  rw [← h.eq, ← sq, ← sq]
  have hc' : Int.gcd x y ≠ 0 := by
    rw [hc]
    exact one_ne_zero
  

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
hc' : Int.gcd x y ≠ 0
⊢ 0 < x ^ 2 + y ^ 2

Next tactic:
cases' Int.ne_zero_of_gcd hc' with hxz hyz
theorem ne_zero_of_coprime (hc : Int.gcd x y = 1) : z ≠ 0 := by
  suffices 0 < z * z by
    rintro rfl
    norm_num at this
  rw [← h.eq, ← sq, ← sq]
  have hc' : Int.gcd x y ≠ 0 := by
    rw [hc]
    exact one_ne_zero
  cases' Int.ne_zero_of_gcd hc' with hxz hyz
  · 

Proof state:
case inl
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
hc' : Int.gcd x y ≠ 0
hxz : x ≠ 0
⊢ 0 < x ^ 2 + y ^ 2

Next tactic:
apply lt_add_of_pos_of_le (sq_pos_of_ne_zero x hxz) (sq_nonneg y)
theorem ne_zero_of_coprime (hc : Int.gcd x y = 1) : z ≠ 0 := by
  suffices 0 < z * z by
    rintro rfl
    norm_num at this
  rw [← h.eq, ← sq, ← sq]
  have hc' : Int.gcd x y ≠ 0 := by
    rw [hc]
    exact one_ne_zero
  cases' Int.ne_zero_of_gcd hc' with hxz hyz
  · apply lt_add_of_pos_of_le (sq_pos_of_ne_zero x hxz) (sq_nonneg y)
  · 

Proof state:
case inr
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
hc' : Int.gcd x y ≠ 0
hyz : y ≠ 0
⊢ 0 < x ^ 2 + y ^ 2

Next tactic:
apply lt_add_of_le_of_pos (sq_nonneg x) (sq_pos_of_ne_zero y hyz)
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
hx : x = 0
⊢ IsPrimitiveClassified h

Next tactic:
subst x
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  

Proof state:
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Int.gcd 0 y = 1
⊢ IsPrimitiveClassified h

Next tactic:
change Nat.gcd 0 (Int.natAbs y) = 1 at hc
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  change Nat.gcd 0 (Int.natAbs y) = 1 at hc
  

Proof state:
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Nat.gcd 0 (Int.natAbs y) = 1
⊢ IsPrimitiveClassified h

Next tactic:
rw [Nat.gcd_zero_left (Int.natAbs y)] at hc
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  change Nat.gcd 0 (Int.natAbs y) = 1 at hc
  rw [Nat.gcd_zero_left (Int.natAbs y)] at hc
  

Proof state:
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Int.natAbs y = 1
⊢ IsPrimitiveClassified h

Next tactic:
cases' Int.natAbs_eq y with hy hy
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  change Nat.gcd 0 (Int.natAbs y) = 1 at hc
  rw [Nat.gcd_zero_left (Int.natAbs y)] at hc
  cases' Int.natAbs_eq y with hy hy
  · 

Proof state:
case inl
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Int.natAbs y = 1
hy : y = ↑(Int.natAbs y)
⊢ IsPrimitiveClassified h

Next tactic:
use 1, 0
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  change Nat.gcd 0 (Int.natAbs y) = 1 at hc
  rw [Nat.gcd_zero_left (Int.natAbs y)] at hc
  cases' Int.natAbs_eq y with hy hy
  · use 1, 0
    

Proof state:
case h
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Int.natAbs y = 1
hy : y = ↑(Int.natAbs y)
⊢ (0 = 1 ^ 2 - 0 ^ 2 ∧ y = 2 * 1 * 0 ∨ 0 = 2 * 1 * 0 ∧ y = 1 ^ 2 - 0 ^ 2) ∧
    Int.gcd 1 0 = 1 ∧ (1 % 2 = 0 ∧ 0 % 2 = 1 ∨ 1 % 2 = 1 ∧ 0 % 2 = 0)

Next tactic:
rw [hy, hc, Int.gcd_zero_right]
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  change Nat.gcd 0 (Int.natAbs y) = 1 at hc
  rw [Nat.gcd_zero_left (Int.natAbs y)] at hc
  cases' Int.natAbs_eq y with hy hy
  · use 1, 0
    rw [hy, hc, Int.gcd_zero_right]
    

Proof state:
case h
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Int.natAbs y = 1
hy : y = ↑(Int.natAbs y)
⊢ (0 = 1 ^ 2 - 0 ^ 2 ∧ ↑1 = 2 * 1 * 0 ∨ 0 = 2 * 1 * 0 ∧ ↑1 = 1 ^ 2 - 0 ^ 2) ∧
    Int.natAbs 1 = 1 ∧ (1 % 2 = 0 ∧ 0 % 2 = 1 ∨ 1 % 2 = 1 ∧ 0 % 2 = 0)

Next tactic:
decide
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  change Nat.gcd 0 (Int.natAbs y) = 1 at hc
  rw [Nat.gcd_zero_left (Int.natAbs y)] at hc
  cases' Int.natAbs_eq y with hy hy
  · use 1, 0
    rw [hy, hc, Int.gcd_zero_right]
    decide
  · 

Proof state:
case inr
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Int.natAbs y = 1
hy : y = -↑(Int.natAbs y)
⊢ IsPrimitiveClassified h

Next tactic:
use 0, 1
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  change Nat.gcd 0 (Int.natAbs y) = 1 at hc
  rw [Nat.gcd_zero_left (Int.natAbs y)] at hc
  cases' Int.natAbs_eq y with hy hy
  · use 1, 0
    rw [hy, hc, Int.gcd_zero_right]
    decide
  · use 0, 1
    

Proof state:
case h
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Int.natAbs y = 1
hy : y = -↑(Int.natAbs y)
⊢ (0 = 0 ^ 2 - 1 ^ 2 ∧ y = 2 * 0 * 1 ∨ 0 = 2 * 0 * 1 ∧ y = 0 ^ 2 - 1 ^ 2) ∧
    Int.gcd 0 1 = 1 ∧ (0 % 2 = 0 ∧ 1 % 2 = 1 ∨ 0 % 2 = 1 ∧ 1 % 2 = 0)

Next tactic:
rw [hy, hc, Int.gcd_zero_left]
theorem isPrimitiveClassified_of_coprime_of_zero_left (hc : Int.gcd x y = 1) (hx : x = 0) :
    h.IsPrimitiveClassified := by
  subst x
  change Nat.gcd 0 (Int.natAbs y) = 1 at hc
  rw [Nat.gcd_zero_left (Int.natAbs y)] at hc
  cases' Int.natAbs_eq y with hy hy
  · use 1, 0
    rw [hy, hc, Int.gcd_zero_right]
    decide
  · use 0, 1
    rw [hy, hc, Int.gcd_zero_left]
    

Proof state:
case h
y z : ℤ
h : PythagoreanTriple 0 y z
hc : Int.natAbs y = 1
hy : y = -↑(Int.natAbs y)
⊢ (0 = 0 ^ 2 - 1 ^ 2 ∧ -↑1 = 2 * 0 * 1 ∨ 0 = 2 * 0 * 1 ∧ -↑1 = 0 ^ 2 - 1 ^ 2) ∧
    Int.natAbs 1 = 1 ∧ (0 % 2 = 0 ∧ 1 % 2 = 1 ∨ 0 % 2 = 1 ∧ 1 % 2 = 0)

Next tactic:
decide
theorem coprime_of_coprime (hc : Int.gcd x y = 1) : Int.gcd y z = 1 := by
  

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
⊢ Int.gcd y z = 1

Next tactic:
by_contra H
theorem coprime_of_coprime (hc : Int.gcd x y = 1) : Int.gcd y z = 1 := by
  by_contra H
  

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
H : ¬Int.gcd y z = 1
⊢ False

Next tactic:
obtain ⟨p, hp, hpy, hpz⟩ := Nat.Prime.not_coprime_iff_dvd.mp H
theorem coprime_of_coprime (hc : Int.gcd x y = 1) : Int.gcd y z = 1 := by
  by_contra H
  obtain ⟨p, hp, hpy, hpz⟩ := Nat.Prime.not_coprime_iff_dvd.mp H
  

Proof state:
case intro.intro.intro
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
H : ¬Int.gcd y z = 1
p : ℕ
hp : Nat.Prime p
hpy : p ∣ Int.natAbs y
hpz : p ∣ Int.natAbs z
⊢ False

Next tactic:
apply hp.not_dvd_one
theorem coprime_of_coprime (hc : Int.gcd x y = 1) : Int.gcd y z = 1 := by
  by_contra H
  obtain ⟨p, hp, hpy, hpz⟩ := Nat.Prime.not_coprime_iff_dvd.mp H
  apply hp.not_dvd_one
  

Proof state:
case intro.intro.intro
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
H : ¬Int.gcd y z = 1
p : ℕ
hp : Nat.Prime p
hpy : p ∣ Int.natAbs y
hpz : p ∣ Int.natAbs z
⊢ p ∣ 1

Next tactic:
rw [← hc]
theorem coprime_of_coprime (hc : Int.gcd x y = 1) : Int.gcd y z = 1 := by
  by_contra H
  obtain ⟨p, hp, hpy, hpz⟩ := Nat.Prime.not_coprime_iff_dvd.mp H
  apply hp.not_dvd_one
  rw [← hc]
  

Proof state:
case intro.intro.intro
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
H : ¬Int.gcd y z = 1
p : ℕ
hp : Nat.Prime p
hpy : p ∣ Int.natAbs y
hpz : p ∣ Int.natAbs z
⊢ p ∣ Int.gcd x y

Next tactic:
apply Nat.dvd_gcd (Int.Prime.dvd_natAbs_of_coe_dvd_sq hp _ _) hpy
theorem coprime_of_coprime (hc : Int.gcd x y = 1) : Int.gcd y z = 1 := by
  by_contra H
  obtain ⟨p, hp, hpy, hpz⟩ := Nat.Prime.not_coprime_iff_dvd.mp H
  apply hp.not_dvd_one
  rw [← hc]
  apply Nat.dvd_gcd (Int.Prime.dvd_natAbs_of_coe_dvd_sq hp _ _) hpy
  

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
H : ¬Int.gcd y z = 1
p : ℕ
hp : Nat.Prime p
hpy : p ∣ Int.natAbs y
hpz : p ∣ Int.natAbs z
⊢ ↑p ∣ x ^ 2

Next tactic:
rw [sq, eq_sub_of_add_eq h]
theorem coprime_of_coprime (hc : Int.gcd x y = 1) : Int.gcd y z = 1 := by
  by_contra H
  obtain ⟨p, hp, hpy, hpz⟩ := Nat.Prime.not_coprime_iff_dvd.mp H
  apply hp.not_dvd_one
  rw [← hc]
  apply Nat.dvd_gcd (Int.Prime.dvd_natAbs_of_coe_dvd_sq hp _ _) hpy
  rw [sq, eq_sub_of_add_eq h]
  

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
H : ¬Int.gcd y z = 1
p : ℕ
hp : Nat.Prime p
hpy : p ∣ Int.natAbs y
hpz : p ∣ Int.natAbs z
⊢ ↑p ∣ z * z - y * y

Next tactic:
rw [← Int.coe_nat_dvd_left] at hpy hpz
theorem coprime_of_coprime (hc : Int.gcd x y = 1) : Int.gcd y z = 1 := by
  by_contra H
  obtain ⟨p, hp, hpy, hpz⟩ := Nat.Prime.not_coprime_iff_dvd.mp H
  apply hp.not_dvd_one
  rw [← hc]
  apply Nat.dvd_gcd (Int.Prime.dvd_natAbs_of_coe_dvd_sq hp _ _) hpy
  rw [sq, eq_sub_of_add_eq h]
  rw [← Int.coe_nat_dvd_left] at hpy hpz
  

Proof state:
x y z : ℤ
h : PythagoreanTriple x y z
hc : Int.gcd x y = 1
H : ¬Int.gcd y z = 1
p : ℕ
hp : Nat.Prime p
hpy : ↑p ∣ y
hpz : ↑p ∣ z
⊢ ↑p ∣ z * z - y * y

Next tactic:
exact dvd_sub (hpz.mul_right _) (hpy.mul_right _)
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
⊢ (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).1 ^ 2 + (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ^ 2 = 1

Next tactic:
field_simp [hk x, div_pow]
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
⊢ (2 * x) ^ 2 + (1 - x ^ 2) ^ 2 = (1 + x ^ 2) ^ 2

Next tactic:
ring
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
⊢ (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ≠ -1

Next tactic:
simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
⊢ ¬1 = -1

Next tactic:
simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1⟩
  invFun p := (p : K × K).1 / ((p : K × K).2 + 1)
  left_inv x := by
    

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
⊢ (fun p => (↑p).1 / ((↑p).2 + 1))
      ((fun x =>
          { val := (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)),
            property :=
              (_ :
                (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).1 ^ 2 +
                      (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ^ 2 =
                    1 ∧
                  (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ≠ -1) })
        x) =
    x

Next tactic:
have h2 : (1 + 1 : K) = 2 := by norm_num
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1⟩
  invFun p := (p : K × K).1 / ((p : K × K).2 + 1)
  left_inv x := by
    have h2 : (1 + 1 : K) = 2 := by 

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
⊢ 1 + 1 = 2

Next tactic:
norm_num
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1⟩
  invFun p := (p : K × K).1 / ((p : K × K).2 + 1)
  left_inv x := by
    have h2 : (1 + 1 : K) = 2 := by norm_num -- Porting note: rfl is not enough to close this
    

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
h2 : 1 + 1 = 2
⊢ (fun p => (↑p).1 / ((↑p).2 + 1))
      ((fun x =>
          { val := (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)),
            property :=
              (_ :
                (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).1 ^ 2 +
                      (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ^ 2 =
                    1 ∧
                  (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ≠ -1) })
        x) =
    x

Next tactic:
have h3 : (2 : K) ≠ 0 := by
      convert hk 1
      rw [one_pow 2, h2]
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1⟩
  invFun p := (p : K × K).1 / ((p : K × K).2 + 1)
  left_inv x := by
    have h2 : (1 + 1 : K) = 2 := by norm_num -- Porting note: rfl is not enough to close this
    have h3 : (2 : K) ≠ 0 := by
      

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
h2 : 1 + 1 = 2
⊢ 2 ≠ 0

Next tactic:
convert hk 1
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1⟩
  invFun p := (p : K × K).1 / ((p : K × K).2 + 1)
  left_inv x := by
    have h2 : (1 + 1 : K) = 2 := by norm_num -- Porting note: rfl is not enough to close this
    have h3 : (2 : K) ≠ 0 := by
      convert hk 1
      

Proof state:
case h.e'_2
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
h2 : 1 + 1 = 2
⊢ 2 = 1 + 1 ^ 2

Next tactic:
rw [one_pow 2, h2]
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1⟩
  invFun p := (p : K × K).1 / ((p : K × K).2 + 1)
  left_inv x := by
    have h2 : (1 + 1 : K) = 2 := by norm_num -- Porting note: rfl is not enough to close this
    have h3 : (2 : K) ≠ 0 := by
      convert hk 1
      rw [one_pow 2, h2]
    

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x : K
h2 : 1 + 1 = 2
h3 : 2 ≠ 0
⊢ (fun p => (↑p).1 / ((↑p).2 + 1))
      ((fun x =>
          { val := (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)),
            property :=
              (_ :
                (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).1 ^ 2 +
                      (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ^ 2 =
                    1 ∧
                  (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ≠ -1) })
        x) =
    x

Next tactic:
field_simp [hk x, h2, add_assoc, add_comm, add_sub_cancel'_right, mul_comm]
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1⟩
  invFun p := (p : K × K).1 / ((p : K × K).2 + 1)
  left_inv x := by
    have h2 : (1 + 1 : K) = 2 := by norm_num -- Porting note: rfl is not enough to close this
    have h3 : (2 : K) ≠ 0 := by
      convert hk 1
      rw [one_pow 2, h2]
    field_simp [hk x, h2, add_assoc, add_comm, add_sub_cancel'_right, mul_comm]
  right_inv := fun ⟨⟨x, y⟩, hxy, hy⟩ => by
    

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x✝ : { p // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 }
x y : K
hxy : (x, y).1 ^ 2 + (x, y).2 ^ 2 = 1
hy : (x, y).2 ≠ -1
⊢ (fun x =>
        { val := (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)),
          property :=
            (_ :
              (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).1 ^ 2 +
                    (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ^ 2 =
                  1 ∧
                (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ≠ -1) })
      ((fun p => (↑p).1 / ((↑p).2 + 1))
        { val := (x, y), property := (_ : (x, y).1 ^ 2 + (x, y).2 ^ 2 = 1 ∧ (x, y).2 ≠ -1) }) =
    { val := (x, y), property := (_ : (x, y).1 ^ 2 + (x, y).2 ^ 2 = 1 ∧ (x, y).2 ≠ -1) }

Next tactic:
change x ^ 2 + y ^ 2 = 1 at hxy
/-- A parameterization of the unit circle that is useful for classifying Pythagorean triples.
 (To be applied in the case where `K = ℚ`.) -/
def circleEquivGen (hk : ∀ x : K, 1 + x ^ 2 ≠ 0) :
    K ≃ { p : K × K // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 } where
  toFun x :=
    ⟨⟨2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)⟩, by
      field_simp [hk x, div_pow]
      ring, by
      simp only [Ne.def, div_eq_iff (hk x), neg_mul, one_mul, neg_add, sub_eq_add_neg, add_left_inj]
      simpa only [eq_neg_iff_add_eq_zero, one_pow] using hk 1⟩
  invFun p := (p : K × K).1 / ((p : K × K).2 + 1)
  left_inv x := by
    have h2 : (1 + 1 : K) = 2 := by norm_num -- Porting note: rfl is not enough to close this
    have h3 : (2 : K) ≠ 0 := by
      convert hk 1
      rw [one_pow 2, h2]
    field_simp [hk x, h2, add_assoc, add_comm, add_sub_cancel'_right, mul_comm]
  right_inv := fun ⟨⟨x, y⟩, hxy, hy⟩ => by
    change x ^ 2 + y ^ 2 = 1 at hxy
    

Proof state:
K : Type u_1
inst✝ : Field K
hk : ∀ (x : K), 1 + x ^ 2 ≠ 0
x✝ : { p // p.1 ^ 2 + p.2 ^ 2 = 1 ∧ p.2 ≠ -1 }
x y : K
hy : (x, y).2 ≠ -1
hxy : x ^ 2 + y ^ 2 = 1
⊢ (fun x =>
        { val := (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)),
          property :=
            (_ :
              (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).1 ^ 2 +
                    (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ^ 2 =
                  1 ∧
                (2 * x / (1 + x ^ 2), (1 - x ^ 2) / (1 + x ^ 2)).2 ≠ -1) })
      ((fun p => (↑p).1 / ((↑p).2 + 1))
        { val := (x, y), property := (_ : (x, y).1 ^ 2 + (x, y).2 ^ 2 = 1 ∧ (x, y).2 ≠ -1) }) =
    { val := (x, y), property := (_ : (x, y).1 ^ 2 + (x, y).2 ^ 2 = 1 ∧ (x, y).2 ≠ -1) }

Next tactic:
have h2 : y + 1 ≠ 0 := mt eq_neg_of_add_eq_zero_left hy