l3lab/ntp-mathlib huggingface dataset:
/-- **Divergence theorem** for functions on the plane along rectangles. It is formulated in terms of
two functions `f g : ℝ × ℝ → E` and an integral over `Icc a b = [a.1, b.1] × [a.2, b.2]`, where
`a b : ℝ × ℝ`, `a ≤ b`. When thinking of `f` and `g` as the two coordinates of a single function
`F : ℝ × ℝ → E × E` and when `E = ℝ`, this is the usual statement that the integral of the
divergence of `F` inside the rectangle equals the integral of the normal derivative of `F` along the
boundary.

See also `MeasureTheory.integral2_divergence_prod_of_hasFDerivWithinAt_off_countable` for a
version that does not assume `a ≤ b` and uses iterated interval integral instead of the integral
over `Icc a b`. -/
theorem integral_divergence_prod_Icc_of_hasFDerivWithinAt_off_countable_of_le (f g : ℝ × ℝ → E)
    (f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E) (a b : ℝ × ℝ) (hle : a ≤ b) (s : Set (ℝ × ℝ)) (hs : s.Countable)
    (Hcf : ContinuousOn f (Icc a b)) (Hcg : ContinuousOn g (Icc a b))
    (Hdf : ∀ x ∈ Ioo a.1 b.1 ×ˢ Ioo a.2 b.2 \ s, HasFDerivAt f (f' x) x)
    (Hdg : ∀ x ∈ Ioo a.1 b.1 ×ˢ Ioo a.2 b.2 \ s, HasFDerivAt g (g' x) x)
    (Hi : IntegrableOn (fun x => f' x (1, 0) + g' x (0, 1)) (Icc a b)) :
    (∫ x in Icc a b, f' x (1, 0) + g' x (0, 1)) =
      (((∫ x in a.1..b.1, g (x, b.2)) - ∫ x in a.1..b.1, g (x, a.2)) +
          ∫ y in a.2..b.2, f (b.1, y)) -
        ∫ y in a.2..b.2, f (a.1, y) :=
  let e : (ℝ × ℝ) ≃L[ℝ] ℝ² := (ContinuousLinearEquiv.finTwoArrow ℝ ℝ).symm
  calc
    (∫ x in Icc a b, f' x (1, 0) + g' x (0, 1)) =
        ∑ i : Fin 2,
          ((∫ x in Icc (e a ∘ i.succAbove) (e b ∘ i.succAbove),
              ![f, g] i (e.symm <| i.insertNth (e b i) x)) -
            ∫ x in Icc (e a ∘ i.succAbove) (e b ∘ i.succAbove),
              ![f, g] i (e.symm <| i.insertNth (e a i) x)) := by
      refine' integral_divergence_of_hasFDerivWithinAt_off_countable_of_equiv e _ _ ![f, g]
        ![f', g'] s hs a b hle _ (fun x hx => _) _ _ Hi
      · exact fun x y => (OrderIso.finTwoArrowIso ℝ).symm.le_iff_le
      · exact (volume_preserving_finTwoArrow ℝ).symm _
      · exact Fin.forall_fin_two.2 ⟨Hcf, Hcg⟩
      · rw [Icc_prod_eq, interior_prod_eq, interior_Icc, interior_Icc] at hx
        exact Fin.forall_fin_two.2 ⟨Hdf x hx, Hdg x hx⟩
      · intro x; rw [Fin.sum_univ_two]; 

Proof state:
case refine'_5
E : Type u
inst✝² : NormedAddCommGroup E
inst✝¹ : NormedSpace ℝ E
inst✝ : CompleteSpace E
f g : ℝ × ℝ → E
f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E
a b : ℝ × ℝ
hle : a ≤ b
s : Set (ℝ × ℝ)
hs : Set.Countable s
Hcf : ContinuousOn f (Set.Icc a b)
Hcg : ContinuousOn g (Set.Icc a b)
Hdf : ∀ x ∈ Set.Ioo a.1 b.1 ×ˢ Set.Ioo a.2 b.2 \ s, HasFDerivAt f (f' x) x
Hdg : ∀ x ∈ Set.Ioo a.1 b.1 ×ˢ Set.Ioo a.2 b.2 \ s, HasFDerivAt g (g' x) x
Hi : IntegrableOn (fun x => (f' x) (1, 0) + (g' x) (0, 1)) (Set.Icc a b)
e : (ℝ × ℝ) ≃L[ℝ] Fin 2 → ℝ := ContinuousLinearEquiv.symm (ContinuousLinearEquiv.finTwoArrow ℝ ℝ)
x : ℝ × ℝ
⊢ (f' x) (1, 0) + (g' x) (0, 1) =
    (Matrix.vecCons f' ![g'] 0 x) ((ContinuousLinearEquiv.symm e) (Pi.single 0 1)) +
      (Matrix.vecCons f' ![g'] 1 x) ((ContinuousLinearEquiv.symm e) (Pi.single 1 1))

Next tactic:
rfl
/-- **Divergence theorem** for functions on the plane. It is formulated in terms of two functions
`f g : ℝ × ℝ → E` and iterated integral `∫ x in a₁..b₁, ∫ y in a₂..b₂, _`, where
`a₁ a₂ b₁ b₂ : ℝ`. When thinking of `f` and `g` as the two coordinates of a single function
`F : ℝ × ℝ → E × E` and when `E = ℝ`, this is the usual statement that the integral of the
divergence of `F` inside the rectangle with vertices `(aᵢ, bⱼ)`, `i, j =1,2`, equals the integral of
the normal derivative of `F` along the boundary.

See also `MeasureTheory.integral_divergence_prod_Icc_of_hasFDerivWithinAt_off_countable_of_le`
for a version that uses an integral over `Icc a b`, where `a b : ℝ × ℝ`, `a ≤ b`. -/
theorem integral2_divergence_prod_of_hasFDerivWithinAt_off_countable (f g : ℝ × ℝ → E)
    (f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E) (a₁ a₂ b₁ b₂ : ℝ) (s : Set (ℝ × ℝ)) (hs : s.Countable)
    (Hcf : ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]]))
    (Hcg : ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]]))
    (Hdf : ∀ x ∈ Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Ioo (min a₂ b₂) (max a₂ b₂) \ s,
      HasFDerivAt f (f' x) x)
    (Hdg : ∀ x ∈ Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Ioo (min a₂ b₂) (max a₂ b₂) \ s,
      HasFDerivAt g (g' x) x)
    (Hi : IntegrableOn (fun x => f' x (1, 0) + g' x (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]])) :
    (∫ x in a₁..b₁, ∫ y in a₂..b₂, f' (x, y) (1, 0) + g' (x, y) (0, 1)) =
      (((∫ x in a₁..b₁, g (x, b₂)) - ∫ x in a₁..b₁, g (x, a₂)) + ∫ y in a₂..b₂, f (b₁, y)) -
        ∫ y in a₂..b₂, f (a₁, y) := by
  

Proof state:
E : Type u
inst✝² : NormedAddCommGroup E
inst✝¹ : NormedSpace ℝ E
inst✝ : CompleteSpace E
f g : ℝ × ℝ → E
f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E
a₁ a₂ b₁ b₂ : ℝ
s : Set (ℝ × ℝ)
hs : Set.Countable s
Hcf : ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
Hcg : ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
Hdf : ∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt f (f' x) x
Hdg : ∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt g (g' x) x
Hi : IntegrableOn (fun x => (f' x) (1, 0) + (g' x) (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
⊢ ∫ (x : ℝ) in a₁..b₁, ∫ (y : ℝ) in a₂..b₂, (f' (x, y)) (1, 0) + (g' (x, y)) (0, 1) =
    (((∫ (x : ℝ) in a₁..b₁, g (x, b₂)) - ∫ (x : ℝ) in a₁..b₁, g (x, a₂)) + ∫ (y : ℝ) in a₂..b₂, f (b₁, y)) -
      ∫ (y : ℝ) in a₂..b₂, f (a₁, y)

Next tactic:
wlog h₁ : a₁ ≤ b₁ generalizing a₁ b₁
/-- **Divergence theorem** for functions on the plane. It is formulated in terms of two functions
`f g : ℝ × ℝ → E` and iterated integral `∫ x in a₁..b₁, ∫ y in a₂..b₂, _`, where
`a₁ a₂ b₁ b₂ : ℝ`. When thinking of `f` and `g` as the two coordinates of a single function
`F : ℝ × ℝ → E × E` and when `E = ℝ`, this is the usual statement that the integral of the
divergence of `F` inside the rectangle with vertices `(aᵢ, bⱼ)`, `i, j =1,2`, equals the integral of
the normal derivative of `F` along the boundary.

See also `MeasureTheory.integral_divergence_prod_Icc_of_hasFDerivWithinAt_off_countable_of_le`
for a version that uses an integral over `Icc a b`, where `a b : ℝ × ℝ`, `a ≤ b`. -/
theorem integral2_divergence_prod_of_hasFDerivWithinAt_off_countable (f g : ℝ × ℝ → E)
    (f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E) (a₁ a₂ b₁ b₂ : ℝ) (s : Set (ℝ × ℝ)) (hs : s.Countable)
    (Hcf : ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]]))
    (Hcg : ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]]))
    (Hdf : ∀ x ∈ Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Ioo (min a₂ b₂) (max a₂ b₂) \ s,
      HasFDerivAt f (f' x) x)
    (Hdg : ∀ x ∈ Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Ioo (min a₂ b₂) (max a₂ b₂) \ s,
      HasFDerivAt g (g' x) x)
    (Hi : IntegrableOn (fun x => f' x (1, 0) + g' x (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]])) :
    (∫ x in a₁..b₁, ∫ y in a₂..b₂, f' (x, y) (1, 0) + g' (x, y) (0, 1)) =
      (((∫ x in a₁..b₁, g (x, b₂)) - ∫ x in a₁..b₁, g (x, a₂)) + ∫ y in a₂..b₂, f (b₁, y)) -
        ∫ y in a₂..b₂, f (a₁, y) := by
  wlog h₁ : a₁ ≤ b₁ generalizing a₁ b₁
  · 

Proof state:
case inr
E : Type u
inst✝² : NormedAddCommGroup E
inst✝¹ : NormedSpace ℝ E
inst✝ : CompleteSpace E
f g : ℝ × ℝ → E
f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E
a₁ a₂ b₁ b₂ : ℝ
s : Set (ℝ × ℝ)
hs : Set.Countable s
Hcf : ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
Hcg : ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
Hdf : ∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt f (f' x) x
Hdg : ∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt g (g' x) x
Hi : IntegrableOn (fun x => (f' x) (1, 0) + (g' x) (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
this :
  ∀ (a₁ b₁ : ℝ),
    ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]]) →
      ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]]) →
        (∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt f (f' x) x) →
          (∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt g (g' x) x) →
            IntegrableOn (fun x => (f' x) (1, 0) + (g' x) (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]]) →
              a₁ ≤ b₁ →
                ∫ (x : ℝ) in a₁..b₁, ∫ (y : ℝ) in a₂..b₂, (f' (x, y)) (1, 0) + (g' (x, y)) (0, 1) =
                  (((∫ (x : ℝ) in a₁..b₁, g (x, b₂)) - ∫ (x : ℝ) in a₁..b₁, g (x, a₂)) +
                      ∫ (y : ℝ) in a₂..b₂, f (b₁, y)) -
                    ∫ (y : ℝ) in a₂..b₂, f (a₁, y)
h₁ : ¬a₁ ≤ b₁
⊢ ∫ (x : ℝ) in a₁..b₁, ∫ (y : ℝ) in a₂..b₂, (f' (x, y)) (1, 0) + (g' (x, y)) (0, 1) =
    (((∫ (x : ℝ) in a₁..b₁, g (x, b₂)) - ∫ (x : ℝ) in a₁..b₁, g (x, a₂)) + ∫ (y : ℝ) in a₂..b₂, f (b₁, y)) -
      ∫ (y : ℝ) in a₂..b₂, f (a₁, y)

Next tactic:
specialize this b₁ a₁
/-- **Divergence theorem** for functions on the plane. It is formulated in terms of two functions
`f g : ℝ × ℝ → E` and iterated integral `∫ x in a₁..b₁, ∫ y in a₂..b₂, _`, where
`a₁ a₂ b₁ b₂ : ℝ`. When thinking of `f` and `g` as the two coordinates of a single function
`F : ℝ × ℝ → E × E` and when `E = ℝ`, this is the usual statement that the integral of the
divergence of `F` inside the rectangle with vertices `(aᵢ, bⱼ)`, `i, j =1,2`, equals the integral of
the normal derivative of `F` along the boundary.

See also `MeasureTheory.integral_divergence_prod_Icc_of_hasFDerivWithinAt_off_countable_of_le`
for a version that uses an integral over `Icc a b`, where `a b : ℝ × ℝ`, `a ≤ b`. -/
theorem integral2_divergence_prod_of_hasFDerivWithinAt_off_countable (f g : ℝ × ℝ → E)
    (f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E) (a₁ a₂ b₁ b₂ : ℝ) (s : Set (ℝ × ℝ)) (hs : s.Countable)
    (Hcf : ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]]))
    (Hcg : ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]]))
    (Hdf : ∀ x ∈ Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Ioo (min a₂ b₂) (max a₂ b₂) \ s,
      HasFDerivAt f (f' x) x)
    (Hdg : ∀ x ∈ Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Ioo (min a₂ b₂) (max a₂ b₂) \ s,
      HasFDerivAt g (g' x) x)
    (Hi : IntegrableOn (fun x => f' x (1, 0) + g' x (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]])) :
    (∫ x in a₁..b₁, ∫ y in a₂..b₂, f' (x, y) (1, 0) + g' (x, y) (0, 1)) =
      (((∫ x in a₁..b₁, g (x, b₂)) - ∫ x in a₁..b₁, g (x, a₂)) + ∫ y in a₂..b₂, f (b₁, y)) -
        ∫ y in a₂..b₂, f (a₁, y) := by
  wlog h₁ : a₁ ≤ b₁ generalizing a₁ b₁
  · specialize this b₁ a₁
    rw [uIcc_comm b₁ a₁, min_comm b₁ a₁, max_comm b₁ a₁] at this
    simp only [intervalIntegral.integral_symm b₁ a₁]
    refine' (congr_arg Neg.neg (this Hcf Hcg Hdf Hdg Hi (le_of_not_le h₁))).trans _; abel
  

Proof state:
E : Type u
inst✝² : NormedAddCommGroup E
inst✝¹ : NormedSpace ℝ E
inst✝ : CompleteSpace E
f g : ℝ × ℝ → E
f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E
a₂ b₂ : ℝ
s : Set (ℝ × ℝ)
hs : Set.Countable s
a₁ b₁ : ℝ
Hcf : ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
Hcg : ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
Hdf : ∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt f (f' x) x
Hdg : ∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt g (g' x) x
Hi : IntegrableOn (fun x => (f' x) (1, 0) + (g' x) (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
h₁ : a₁ ≤ b₁
⊢ ∫ (x : ℝ) in a₁..b₁, ∫ (y : ℝ) in a₂..b₂, (f' (x, y)) (1, 0) + (g' (x, y)) (0, 1) =
    (((∫ (x : ℝ) in a₁..b₁, g (x, b₂)) - ∫ (x : ℝ) in a₁..b₁, g (x, a₂)) + ∫ (y : ℝ) in a₂..b₂, f (b₁, y)) -
      ∫ (y : ℝ) in a₂..b₂, f (a₁, y)

Next tactic:
wlog h₂ : a₂ ≤ b₂ generalizing a₂ b₂
/-- **Divergence theorem** for functions on the plane. It is formulated in terms of two functions
`f g : ℝ × ℝ → E` and iterated integral `∫ x in a₁..b₁, ∫ y in a₂..b₂, _`, where
`a₁ a₂ b₁ b₂ : ℝ`. When thinking of `f` and `g` as the two coordinates of a single function
`F : ℝ × ℝ → E × E` and when `E = ℝ`, this is the usual statement that the integral of the
divergence of `F` inside the rectangle with vertices `(aᵢ, bⱼ)`, `i, j =1,2`, equals the integral of
the normal derivative of `F` along the boundary.

See also `MeasureTheory.integral_divergence_prod_Icc_of_hasFDerivWithinAt_off_countable_of_le`
for a version that uses an integral over `Icc a b`, where `a b : ℝ × ℝ`, `a ≤ b`. -/
theorem integral2_divergence_prod_of_hasFDerivWithinAt_off_countable (f g : ℝ × ℝ → E)
    (f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E) (a₁ a₂ b₁ b₂ : ℝ) (s : Set (ℝ × ℝ)) (hs : s.Countable)
    (Hcf : ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]]))
    (Hcg : ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]]))
    (Hdf : ∀ x ∈ Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Ioo (min a₂ b₂) (max a₂ b₂) \ s,
      HasFDerivAt f (f' x) x)
    (Hdg : ∀ x ∈ Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Ioo (min a₂ b₂) (max a₂ b₂) \ s,
      HasFDerivAt g (g' x) x)
    (Hi : IntegrableOn (fun x => f' x (1, 0) + g' x (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]])) :
    (∫ x in a₁..b₁, ∫ y in a₂..b₂, f' (x, y) (1, 0) + g' (x, y) (0, 1)) =
      (((∫ x in a₁..b₁, g (x, b₂)) - ∫ x in a₁..b₁, g (x, a₂)) + ∫ y in a₂..b₂, f (b₁, y)) -
        ∫ y in a₂..b₂, f (a₁, y) := by
  wlog h₁ : a₁ ≤ b₁ generalizing a₁ b₁
  · specialize this b₁ a₁
    rw [uIcc_comm b₁ a₁, min_comm b₁ a₁, max_comm b₁ a₁] at this
    simp only [intervalIntegral.integral_symm b₁ a₁]
    refine' (congr_arg Neg.neg (this Hcf Hcg Hdf Hdg Hi (le_of_not_le h₁))).trans _; abel
  wlog h₂ : a₂ ≤ b₂ generalizing a₂ b₂
  · specialize this b₂ a₂
    rw [uIcc_comm b₂ a₂, min_comm b₂ a₂, max_comm b₂ a₂] at this
    simp only [intervalIntegral.integral_symm b₂ a₂, intervalIntegral.integral_neg]
    refine' (congr_arg Neg.neg (this Hcf Hcg Hdf Hdg Hi (le_of_not_le h₂))).trans _; abel
  

Proof state:
E : Type u
inst✝² : NormedAddCommGroup E
inst✝¹ : NormedSpace ℝ E
inst✝ : CompleteSpace E
f g : ℝ × ℝ → E
f' g' : ℝ × ℝ → ℝ × ℝ →L[ℝ] E
s : Set (ℝ × ℝ)
hs : Set.Countable s
a₁ b₁ : ℝ
h₁ : a₁ ≤ b₁
a₂ b₂ : ℝ
Hcf : ContinuousOn f ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
Hcg : ContinuousOn g ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
Hdf : ∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt f (f' x) x
Hdg : ∀ x ∈ Set.Ioo (min a₁ b₁) (max a₁ b₁) ×ˢ Set.Ioo (min a₂ b₂) (max a₂ b₂) \ s, HasFDerivAt g (g' x) x
Hi : IntegrableOn (fun x => (f' x) (1, 0) + (g' x) (0, 1)) ([[a₁, b₁]] ×ˢ [[a₂, b₂]])
h₂ : a₂ ≤ b₂
⊢ ∫ (x : ℝ) in a₁..b₁, ∫ (y : ℝ) in a₂..b₂, (f' (x, y)) (1, 0) + (g' (x, y)) (0, 1) =
    (((∫ (x : ℝ) in a₁..b₁, g (x, b₂)) - ∫ (x : ℝ) in a₁..b₁, g (x, a₂)) + ∫ (y : ℝ) in a₂..b₂, f (b₁, y)) -
      ∫ (y : ℝ) in a₂..b₂, f (a₁, y)

Next tactic:
simp only [uIcc_of_le h₁, uIcc_of_le h₂, min_eq_left, max_eq_right, h₁, h₂] at Hcf Hcg Hdf Hdg Hi
/-- If `n` passes the Fermat primality test to base `b`, then `n` is coprime with `b`, assuming that
`n` and `b` are both positive.
-/
theorem coprime_of_probablePrime {n b : ℕ} (h : ProbablePrime n b) (h₁ : 1 ≤ n) (h₂ : 1 ≤ b) :
    Nat.Coprime n b := by
  

Proof state:
n b : ℕ
h : ProbablePrime n b
h₁ : 1 ≤ n
h₂ : 1 ≤ b
⊢ Coprime n b

Next tactic:
by_cases h₃ : 2 ≤ n
/-- If `n` passes the Fermat primality test to base `b`, then `n` is coprime with `b`, assuming that
`n` and `b` are both positive.
-/
theorem coprime_of_probablePrime {n b : ℕ} (h : ProbablePrime n b) (h₁ : 1 ≤ n) (h₂ : 1 ≤ b) :
    Nat.Coprime n b := by
  by_cases h₃ : 2 ≤ n
  · -- To prove that `n` is coprime with `b`, we need to show that for all prime factors of `n`,
    -- we can derive a contradiction if `n` divides `b`.
    

Proof state:
case pos
n b : ℕ
h : ProbablePrime n b
h₁ : 1 ≤ n
h₂ : 1 ≤ b
h₃ : 2 ≤ n
⊢ Coprime n b

Next tactic:
apply Nat.coprime_of_dvd
/-- If `n` passes the Fermat primality test to base `b`, then `n` is coprime with `b`, assuming that
`n` and `b` are both positive.
-/
theorem coprime_of_probablePrime {n b : ℕ} (h : ProbablePrime n b) (h₁ : 1 ≤ n) (h₂ : 1 ≤ b) :
    Nat.Coprime n b := by
  by_cases h₃ : 2 ≤ n
  · -- To prove that `n` is coprime with `b`, we need to show that for all prime factors of `n`,
    -- we can derive a contradiction if `n` divides `b`.
    apply Nat.coprime_of_dvd
    -- If `k` is a prime number that divides both `n` and `b`, then we know that `n = m * k` and
    -- `b = j * k` for some natural numbers `m` and `j`. We substitute these into the hypothesis.
    

Proof state:
case pos.H
n b : ℕ
h : ProbablePrime n b
h₁ : 1 ≤ n
h₂ : 1 ≤ b
h₃ : 2 ≤ n
⊢ ∀ (k : ℕ), Prime k → k ∣ n → ¬k ∣ b

Next tactic:
rintro k hk ⟨m, rfl⟩ ⟨j, rfl⟩
/-- If `n` passes the Fermat primality test to base `b`, then `n` is coprime with `b`, assuming that
`n` and `b` are both positive.
-/
theorem coprime_of_probablePrime {n b : ℕ} (h : ProbablePrime n b) (h₁ : 1 ≤ n) (h₂ : 1 ≤ b) :
    Nat.Coprime n b := by
  by_cases h₃ : 2 ≤ n
  · -- To prove that `n` is coprime with `b`, we need to show that for all prime factors of `n`,
    -- we can derive a contradiction if `n` divides `b`.
    apply Nat.coprime_of_dvd
    -- If `k` is a prime number that divides both `n` and `b`, then we know that `n = m * k` and
    -- `b = j * k` for some natural numbers `m` and `j`. We substitute these into the hypothesis.
    rintro k hk ⟨m, rfl⟩ ⟨j, rfl⟩
    -- Because prime numbers do not divide 1, it suffices to show that `k ∣ 1` to prove a
    -- contradiction
    

Proof state:
case pos.H.intro.intro
k : ℕ
hk : Prime k
m : ℕ
h₁ : 1 ≤ k * m
h₃ : 2 ≤ k * m
j : ℕ
h₂ : 1 ≤ k * j
h : ProbablePrime (k * m) (k * j)
⊢ False

Next tactic:
apply Nat.Prime.not_dvd_one hk
/-- If `n` passes the Fermat primality test to base `b`, then `n` is coprime with `b`, assuming that
`n` and `b` are both positive.
-/
theorem coprime_of_probablePrime {n b : ℕ} (h : ProbablePrime n b) (h₁ : 1 ≤ n) (h₂ : 1 ≤ b) :
    Nat.Coprime n b := by
  by_cases h₃ : 2 ≤ n
  · -- To prove that `n` is coprime with `b`, we need to show that for all prime factors of `n`,
    -- we can derive a contradiction if `n` divides `b`.
    apply Nat.coprime_of_dvd
    -- If `k` is a prime number that divides both `n` and `b`, then we know that `n = m * k` and
    -- `b = j * k` for some natural numbers `m` and `j`. We substitute these into the hypothesis.
    rintro k hk ⟨m, rfl⟩ ⟨j, rfl⟩
    -- Because prime numbers do not divide 1, it suffices to show that `k ∣ 1` to prove a
    -- contradiction
    apply Nat.Prime.not_dvd_one hk
    -- Since `n` divides `b ^ (n - 1) - 1`, `k` also divides `b ^ (n - 1) - 1`
    

Proof state:
case pos.H.intro.intro
k : ℕ
hk : Prime k
m : ℕ
h₁ : 1 ≤ k * m
h₃ : 2 ≤ k * m
j : ℕ
h₂ : 1 ≤ k * j
h : ProbablePrime (k * m) (k * j)
⊢ k ∣ 1

Next tactic:
replace h := dvd_of_mul_right_dvd h
/-- If `n` passes the Fermat primality test to base `b`, then `n` is coprime with `b`, assuming that
`n` and `b` are both positive.
-/
theorem coprime_of_probablePrime {n b : ℕ} (h : ProbablePrime n b) (h₁ : 1 ≤ n) (h₂ : 1 ≤ b) :
    Nat.Coprime n b := by
  by_cases h₃ : 2 ≤ n
  · -- To prove that `n` is coprime with `b`, we need to show that for all prime factors of `n`,
    -- we can derive a contradiction if `n` divides `b`.
    apply Nat.coprime_of_dvd
    -- If `k` is a prime number that divides both `n` and `b`, then we know that `n = m * k` and
    -- `b = j * k` for some natural numbers `m` and `j`. We substitute these into the hypothesis.
    rintro k hk ⟨m, rfl⟩ ⟨j, rfl⟩
    -- Because prime numbers do not divide 1, it suffices to show that `k ∣ 1` to prove a
    -- contradiction
    apply Nat.Prime.not_dvd_one hk
    -- Since `n` divides `b ^ (n - 1) - 1`, `k` also divides `b ^ (n - 1) - 1`
    replace h := dvd_of_mul_right_dvd h
    -- Because `k` divides `b ^ (n - 1) - 1`, if we can show that `k` also divides `b ^ (n - 1)`,
    -- then we know `k` divides 1.
    

Proof state:
case pos.H.intro.intro
k : ℕ
hk : Prime k
m : ℕ
h₁ : 1 ≤ k * m
h₃ : 2 ≤ k * m
j : ℕ
h₂ : 1 ≤ k * j
h : k ∣ (k * j) ^ (k * m - 1) - 1
⊢ k ∣ 1

Next tactic:
rw [Nat.dvd_add_iff_right h, Nat.sub_add_cancel (Nat.one_le_pow _ _ h₂)]
/-- If `n` passes the Fermat primality test to base `b`, then `n` is coprime with `b`, assuming that
`n` and `b` are both positive.
-/
theorem coprime_of_probablePrime {n b : ℕ} (h : ProbablePrime n b) (h₁ : 1 ≤ n) (h₂ : 1 ≤ b) :
    Nat.Coprime n b := by
  by_cases h₃ : 2 ≤ n
  · -- To prove that `n` is coprime with `b`, we need to show that for all prime factors of `n`,
    -- we can derive a contradiction if `n` divides `b`.
    apply Nat.coprime_of_dvd
    -- If `k` is a prime number that divides both `n` and `b`, then we know that `n = m * k` and
    -- `b = j * k` for some natural numbers `m` and `j`. We substitute these into the hypothesis.
    rintro k hk ⟨m, rfl⟩ ⟨j, rfl⟩
    -- Because prime numbers do not divide 1, it suffices to show that `k ∣ 1` to prove a
    -- contradiction
    apply Nat.Prime.not_dvd_one hk
    -- Since `n` divides `b ^ (n - 1) - 1`, `k` also divides `b ^ (n - 1) - 1`
    replace h := dvd_of_mul_right_dvd h
    -- Because `k` divides `b ^ (n - 1) - 1`, if we can show that `k` also divides `b ^ (n - 1)`,
    -- then we know `k` divides 1.
    rw [Nat.dvd_add_iff_right h, Nat.sub_add_cancel (Nat.one_le_pow _ _ h₂)]
    -- Since `k` divides `b`, `k` also divides any power of `b` except `b ^ 0`. Therefore, it
    -- suffices to show that `n - 1` isn't zero. However, we know that `n - 1` isn't zero because we
    -- assumed `2 ≤ n` when doing `by_cases`.
    

Proof state:
case pos.H.intro.intro
k : ℕ
hk : Prime k
m : ℕ
h₁ : 1 ≤ k * m
h₃ : 2 ≤ k * m
j : ℕ
h₂ : 1 ≤ k * j
h : k ∣ (k * j) ^ (k * m - 1) - 1
⊢ k ∣ (k * j) ^ (k * m - 1)

Next tactic:
refine' dvd_of_mul_right_dvd (dvd_pow_self (k * j) _)
/-- If `n` passes the Fermat primality test to base `b`, then `n` is coprime with `b`, assuming that
`n` and `b` are both positive.
-/
theorem coprime_of_probablePrime {n b : ℕ} (h : ProbablePrime n b) (h₁ : 1 ≤ n) (h₂ : 1 ≤ b) :
    Nat.Coprime n b := by
  by_cases h₃ : 2 ≤ n
  · -- To prove that `n` is coprime with `b`, we need to show that for all prime factors of `n`,
    -- we can derive a contradiction if `n` divides `b`.
    apply Nat.coprime_of_dvd
    -- If `k` is a prime number that divides both `n` and `b`, then we know that `n = m * k` and
    -- `b = j * k` for some natural numbers `m` and `j`. We substitute these into the hypothesis.
    rintro k hk ⟨m, rfl⟩ ⟨j, rfl⟩
    -- Because prime numbers do not divide 1, it suffices to show that `k ∣ 1` to prove a
    -- contradiction
    apply Nat.Prime.not_dvd_one hk
    -- Since `n` divides `b ^ (n - 1) - 1`, `k` also divides `b ^ (n - 1) - 1`
    replace h := dvd_of_mul_right_dvd h
    -- Because `k` divides `b ^ (n - 1) - 1`, if we can show that `k` also divides `b ^ (n - 1)`,
    -- then we know `k` divides 1.
    rw [Nat.dvd_add_iff_right h, Nat.sub_add_cancel (Nat.one_le_pow _ _ h₂)]
    -- Since `k` divides `b`, `k` also divides any power of `b` except `b ^ 0`. Therefore, it
    -- suffices to show that `n - 1` isn't zero. However, we know that `n - 1` isn't zero because we
    -- assumed `2 ≤ n` when doing `by_cases`.
    refine' dvd_of_mul_right_dvd (dvd_pow_self (k * j) _)
    

Proof state:
case pos.H.intro.intro
k : ℕ
hk : Prime k
m : ℕ
h₁ : 1 ≤ k * m
h₃ : 2 ≤ k * m
j : ℕ
h₂ : 1 ≤ k * j
h : k ∣ (k * j) ^ (k * m - 1) - 1
⊢ k * m - 1 ≠ 0

Next tactic:
linarith [tsub_pos_of_lt (one_lt_two.trans_le h₃)]