l3lab/ntp-mathlib huggingface dataset:
/-- Consider a function `d : X → X → ℝ≥0` such that `d x x = 0` and `d x y = d y x` for all `x`,
`y`. Let `dist` be the largest pseudometric distance such that `dist x y ≤ d x y`, see
`PseudoMetricSpace.ofPreNNDist`. Suppose that `d` satisfies the following triangle-like
inequality: `d x₁ x₄ ≤ 2 * max (d x₁ x₂, d x₂ x₃, d x₃ x₄)`. Then `d x y ≤ 2 * dist x y` for all
`x`, `y`. -/
theorem le_two_mul_dist_ofPreNNDist (d : X → X → ℝ≥0) (dist_self : ∀ x, d x x = 0)
    (dist_comm : ∀ x y, d x y = d y x)
    (hd : ∀ x₁ x₂ x₃ x₄, d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))) (x y : X) :
    ↑(d x y) ≤ 2 * @dist X
      (@PseudoMetricSpace.toDist X (PseudoMetricSpace.ofPreNNDist d dist_self dist_comm)) x y := by
  /- We need to show that `d x y` is at most twice the sum `L` of `d xᵢ xᵢ₊₁` over a path
    `x₀=x, ..., xₙ=y`. We prove it by induction on the length `n` of the sequence. Find an edge that
    splits the path into two parts of almost equal length: both `d x₀ x₁ + ... + d xₖ₋₁ xₖ` and
    `d xₖ₊₁ xₖ₊₂ + ... + d xₙ₋₁ xₙ` are less than or equal to `L / 2`.
    Then `d x₀ xₖ ≤ L`, `d xₖ xₖ₊₁ ≤ L`, and `d xₖ₊₁ xₙ ≤ L`, thus `d x₀ xₙ ≤ 2 * L`. -/
  rw [dist_ofPreNNDist, ← NNReal.coe_two, ← NNReal.coe_mul, NNReal.mul_iInf, NNReal.coe_le_coe]
  refine' le_ciInf fun l => _
  have hd₀_trans : Transitive fun x y => d x y = 0 := by
    intro a b c hab hbc
    rw [← nonpos_iff_eq_zero]
    simpa only [nonpos_iff_eq_zero, hab, hbc, dist_self c, max_self, mul_zero] using hd a b c c
  haveI : IsTrans X fun x y => d x y = 0 := ⟨hd₀_trans⟩
  induction' hn : length l using Nat.strong_induction_on with n ihn generalizing x y l
  simp only at ihn
  subst n
  set L := zipWith d (x::l) (l ++ [y])
  have hL_len : length L = length l + 1 := by simp
  rcases eq_or_ne (d x y) 0 with hd₀ | hd₀
  · simp only [hd₀, zero_le]
  rsuffices ⟨z, z', hxz, hzz', hz'y⟩ : ∃ z z' : X, d x z ≤ L.sum ∧ d z z' ≤ L.sum ∧ d z' y ≤ L.sum
  · exact (hd x z z' y).trans (mul_le_mul_left' (max_le hxz (max_le hzz' hz'y)) _)
  set s : Set ℕ := { m : ℕ | 2 * (take m L).sum ≤ L.sum }
  

Proof state:
X : Type u_1
d : X → X → ℝ≥0
dist_self : ∀ (x : X), d x x = 0
dist_comm : ∀ (x y : X), d x y = d y x
hd : ∀ (x₁ x₂ x₃ x₄ : X), d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))
hd₀_trans : Transitive fun x y => d x y = 0
this : IsTrans X fun x y => d x y = 0
x y : X
l : List X
ihn : ∀ m < length l, ∀ (x y : X) (l : List X), length l = m → d x y ≤ 2 * sum (zipWith d (x :: l) (l ++ [y]))
L : List ℝ≥0 := zipWith d (x :: l) (l ++ [y])
hL_len : length L = length l + 1
hd₀ : d x y ≠ 0
s : Set ℕ := {m | 2 * sum (take m L) ≤ sum L}
⊢ ∃ z z', d x z ≤ sum L ∧ d z z' ≤ sum L ∧ d z' y ≤ sum L

Next tactic:
have hs₀ : 0 ∈ s := by simp
/-- Consider a function `d : X → X → ℝ≥0` such that `d x x = 0` and `d x y = d y x` for all `x`,
`y`. Let `dist` be the largest pseudometric distance such that `dist x y ≤ d x y`, see
`PseudoMetricSpace.ofPreNNDist`. Suppose that `d` satisfies the following triangle-like
inequality: `d x₁ x₄ ≤ 2 * max (d x₁ x₂, d x₂ x₃, d x₃ x₄)`. Then `d x y ≤ 2 * dist x y` for all
`x`, `y`. -/
theorem le_two_mul_dist_ofPreNNDist (d : X → X → ℝ≥0) (dist_self : ∀ x, d x x = 0)
    (dist_comm : ∀ x y, d x y = d y x)
    (hd : ∀ x₁ x₂ x₃ x₄, d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))) (x y : X) :
    ↑(d x y) ≤ 2 * @dist X
      (@PseudoMetricSpace.toDist X (PseudoMetricSpace.ofPreNNDist d dist_self dist_comm)) x y := by
  /- We need to show that `d x y` is at most twice the sum `L` of `d xᵢ xᵢ₊₁` over a path
    `x₀=x, ..., xₙ=y`. We prove it by induction on the length `n` of the sequence. Find an edge that
    splits the path into two parts of almost equal length: both `d x₀ x₁ + ... + d xₖ₋₁ xₖ` and
    `d xₖ₊₁ xₖ₊₂ + ... + d xₙ₋₁ xₙ` are less than or equal to `L / 2`.
    Then `d x₀ xₖ ≤ L`, `d xₖ xₖ₊₁ ≤ L`, and `d xₖ₊₁ xₙ ≤ L`, thus `d x₀ xₙ ≤ 2 * L`. -/
  rw [dist_ofPreNNDist, ← NNReal.coe_two, ← NNReal.coe_mul, NNReal.mul_iInf, NNReal.coe_le_coe]
  refine' le_ciInf fun l => _
  have hd₀_trans : Transitive fun x y => d x y = 0 := by
    intro a b c hab hbc
    rw [← nonpos_iff_eq_zero]
    simpa only [nonpos_iff_eq_zero, hab, hbc, dist_self c, max_self, mul_zero] using hd a b c c
  haveI : IsTrans X fun x y => d x y = 0 := ⟨hd₀_trans⟩
  induction' hn : length l using Nat.strong_induction_on with n ihn generalizing x y l
  simp only at ihn
  subst n
  set L := zipWith d (x::l) (l ++ [y])
  have hL_len : length L = length l + 1 := by simp
  rcases eq_or_ne (d x y) 0 with hd₀ | hd₀
  · simp only [hd₀, zero_le]
  rsuffices ⟨z, z', hxz, hzz', hz'y⟩ : ∃ z z' : X, d x z ≤ L.sum ∧ d z z' ≤ L.sum ∧ d z' y ≤ L.sum
  · exact (hd x z z' y).trans (mul_le_mul_left' (max_le hxz (max_le hzz' hz'y)) _)
  set s : Set ℕ := { m : ℕ | 2 * (take m L).sum ≤ L.sum }
  have hs₀ : 0 ∈ s := by 

Proof state:
X : Type u_1
d : X → X → ℝ≥0
dist_self : ∀ (x : X), d x x = 0
dist_comm : ∀ (x y : X), d x y = d y x
hd : ∀ (x₁ x₂ x₃ x₄ : X), d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))
hd₀_trans : Transitive fun x y => d x y = 0
this : IsTrans X fun x y => d x y = 0
x y : X
l : List X
ihn : ∀ m < length l, ∀ (x y : X) (l : List X), length l = m → d x y ≤ 2 * sum (zipWith d (x :: l) (l ++ [y]))
L : List ℝ≥0 := zipWith d (x :: l) (l ++ [y])
hL_len : length L = length l + 1
hd₀ : d x y ≠ 0
s : Set ℕ := {m | 2 * sum (take m L) ≤ sum L}
⊢ 0 ∈ s

Next tactic:
simp
/-- Consider a function `d : X → X → ℝ≥0` such that `d x x = 0` and `d x y = d y x` for all `x`,
`y`. Let `dist` be the largest pseudometric distance such that `dist x y ≤ d x y`, see
`PseudoMetricSpace.ofPreNNDist`. Suppose that `d` satisfies the following triangle-like
inequality: `d x₁ x₄ ≤ 2 * max (d x₁ x₂, d x₂ x₃, d x₃ x₄)`. Then `d x y ≤ 2 * dist x y` for all
`x`, `y`. -/
theorem le_two_mul_dist_ofPreNNDist (d : X → X → ℝ≥0) (dist_self : ∀ x, d x x = 0)
    (dist_comm : ∀ x y, d x y = d y x)
    (hd : ∀ x₁ x₂ x₃ x₄, d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))) (x y : X) :
    ↑(d x y) ≤ 2 * @dist X
      (@PseudoMetricSpace.toDist X (PseudoMetricSpace.ofPreNNDist d dist_self dist_comm)) x y := by
  /- We need to show that `d x y` is at most twice the sum `L` of `d xᵢ xᵢ₊₁` over a path
    `x₀=x, ..., xₙ=y`. We prove it by induction on the length `n` of the sequence. Find an edge that
    splits the path into two parts of almost equal length: both `d x₀ x₁ + ... + d xₖ₋₁ xₖ` and
    `d xₖ₊₁ xₖ₊₂ + ... + d xₙ₋₁ xₙ` are less than or equal to `L / 2`.
    Then `d x₀ xₖ ≤ L`, `d xₖ xₖ₊₁ ≤ L`, and `d xₖ₊₁ xₙ ≤ L`, thus `d x₀ xₙ ≤ 2 * L`. -/
  rw [dist_ofPreNNDist, ← NNReal.coe_two, ← NNReal.coe_mul, NNReal.mul_iInf, NNReal.coe_le_coe]
  refine' le_ciInf fun l => _
  have hd₀_trans : Transitive fun x y => d x y = 0 := by
    intro a b c hab hbc
    rw [← nonpos_iff_eq_zero]
    simpa only [nonpos_iff_eq_zero, hab, hbc, dist_self c, max_self, mul_zero] using hd a b c c
  haveI : IsTrans X fun x y => d x y = 0 := ⟨hd₀_trans⟩
  induction' hn : length l using Nat.strong_induction_on with n ihn generalizing x y l
  simp only at ihn
  subst n
  set L := zipWith d (x::l) (l ++ [y])
  have hL_len : length L = length l + 1 := by simp
  rcases eq_or_ne (d x y) 0 with hd₀ | hd₀
  · simp only [hd₀, zero_le]
  rsuffices ⟨z, z', hxz, hzz', hz'y⟩ : ∃ z z' : X, d x z ≤ L.sum ∧ d z z' ≤ L.sum ∧ d z' y ≤ L.sum
  · exact (hd x z z' y).trans (mul_le_mul_left' (max_le hxz (max_le hzz' hz'y)) _)
  set s : Set ℕ := { m : ℕ | 2 * (take m L).sum ≤ L.sum }
  have hs₀ : 0 ∈ s := by simp
  

Proof state:
X : Type u_1
d : X → X → ℝ≥0
dist_self : ∀ (x : X), d x x = 0
dist_comm : ∀ (x y : X), d x y = d y x
hd : ∀ (x₁ x₂ x₃ x₄ : X), d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))
hd₀_trans : Transitive fun x y => d x y = 0
this : IsTrans X fun x y => d x y = 0
x y : X
l : List X
ihn : ∀ m < length l, ∀ (x y : X) (l : List X), length l = m → d x y ≤ 2 * sum (zipWith d (x :: l) (l ++ [y]))
L : List ℝ≥0 := zipWith d (x :: l) (l ++ [y])
hL_len : length L = length l + 1
hd₀ : d x y ≠ 0
s : Set ℕ := {m | 2 * sum (take m L) ≤ sum L}
hs₀ : 0 ∈ s
⊢ ∃ z z', d x z ≤ sum L ∧ d z z' ≤ sum L ∧ d z' y ≤ sum L

Next tactic:
have hsne : s.Nonempty := ⟨0, hs₀⟩
/-- Consider a function `d : X → X → ℝ≥0` such that `d x x = 0` and `d x y = d y x` for all `x`,
`y`. Let `dist` be the largest pseudometric distance such that `dist x y ≤ d x y`, see
`PseudoMetricSpace.ofPreNNDist`. Suppose that `d` satisfies the following triangle-like
inequality: `d x₁ x₄ ≤ 2 * max (d x₁ x₂, d x₂ x₃, d x₃ x₄)`. Then `d x y ≤ 2 * dist x y` for all
`x`, `y`. -/
theorem le_two_mul_dist_ofPreNNDist (d : X → X → ℝ≥0) (dist_self : ∀ x, d x x = 0)
    (dist_comm : ∀ x y, d x y = d y x)
    (hd : ∀ x₁ x₂ x₃ x₄, d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))) (x y : X) :
    ↑(d x y) ≤ 2 * @dist X
      (@PseudoMetricSpace.toDist X (PseudoMetricSpace.ofPreNNDist d dist_self dist_comm)) x y := by
  /- We need to show that `d x y` is at most twice the sum `L` of `d xᵢ xᵢ₊₁` over a path
    `x₀=x, ..., xₙ=y`. We prove it by induction on the length `n` of the sequence. Find an edge that
    splits the path into two parts of almost equal length: both `d x₀ x₁ + ... + d xₖ₋₁ xₖ` and
    `d xₖ₊₁ xₖ₊₂ + ... + d xₙ₋₁ xₙ` are less than or equal to `L / 2`.
    Then `d x₀ xₖ ≤ L`, `d xₖ xₖ₊₁ ≤ L`, and `d xₖ₊₁ xₙ ≤ L`, thus `d x₀ xₙ ≤ 2 * L`. -/
  rw [dist_ofPreNNDist, ← NNReal.coe_two, ← NNReal.coe_mul, NNReal.mul_iInf, NNReal.coe_le_coe]
  refine' le_ciInf fun l => _
  have hd₀_trans : Transitive fun x y => d x y = 0 := by
    intro a b c hab hbc
    rw [← nonpos_iff_eq_zero]
    simpa only [nonpos_iff_eq_zero, hab, hbc, dist_self c, max_self, mul_zero] using hd a b c c
  haveI : IsTrans X fun x y => d x y = 0 := ⟨hd₀_trans⟩
  induction' hn : length l using Nat.strong_induction_on with n ihn generalizing x y l
  simp only at ihn
  subst n
  set L := zipWith d (x::l) (l ++ [y])
  have hL_len : length L = length l + 1 := by simp
  rcases eq_or_ne (d x y) 0 with hd₀ | hd₀
  · simp only [hd₀, zero_le]
  rsuffices ⟨z, z', hxz, hzz', hz'y⟩ : ∃ z z' : X, d x z ≤ L.sum ∧ d z z' ≤ L.sum ∧ d z' y ≤ L.sum
  · exact (hd x z z' y).trans (mul_le_mul_left' (max_le hxz (max_le hzz' hz'y)) _)
  set s : Set ℕ := { m : ℕ | 2 * (take m L).sum ≤ L.sum }
  have hs₀ : 0 ∈ s := by simp
  have hsne : s.Nonempty := ⟨0, hs₀⟩
  obtain ⟨M, hMl, hMs⟩ : ∃ M ≤ length l, IsGreatest s M := by
    have hs_ub : length l ∈ upperBounds s := by
      

Proof state:
X : Type u_1
d : X → X → ℝ≥0
dist_self : ∀ (x : X), d x x = 0
dist_comm : ∀ (x y : X), d x y = d y x
hd : ∀ (x₁ x₂ x₃ x₄ : X), d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))
hd₀_trans : Transitive fun x y => d x y = 0
this : IsTrans X fun x y => d x y = 0
x y : X
l : List X
ihn : ∀ m < length l, ∀ (x y : X) (l : List X), length l = m → d x y ≤ 2 * sum (zipWith d (x :: l) (l ++ [y]))
L : List ℝ≥0 := zipWith d (x :: l) (l ++ [y])
hL_len : length L = length l + 1
hd₀ : d x y ≠ 0
s : Set ℕ := {m | 2 * sum (take m L) ≤ sum L}
hs₀ : 0 ∈ s
hsne : Set.Nonempty s
⊢ length l ∈ upperBounds s

Next tactic:
intro m hm
/-- Consider a function `d : X → X → ℝ≥0` such that `d x x = 0` and `d x y = d y x` for all `x`,
`y`. Let `dist` be the largest pseudometric distance such that `dist x y ≤ d x y`, see
`PseudoMetricSpace.ofPreNNDist`. Suppose that `d` satisfies the following triangle-like
inequality: `d x₁ x₄ ≤ 2 * max (d x₁ x₂, d x₂ x₃, d x₃ x₄)`. Then `d x y ≤ 2 * dist x y` for all
`x`, `y`. -/
theorem le_two_mul_dist_ofPreNNDist (d : X → X → ℝ≥0) (dist_self : ∀ x, d x x = 0)
    (dist_comm : ∀ x y, d x y = d y x)
    (hd : ∀ x₁ x₂ x₃ x₄, d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))) (x y : X) :
    ↑(d x y) ≤ 2 * @dist X
      (@PseudoMetricSpace.toDist X (PseudoMetricSpace.ofPreNNDist d dist_self dist_comm)) x y := by
  /- We need to show that `d x y` is at most twice the sum `L` of `d xᵢ xᵢ₊₁` over a path
    `x₀=x, ..., xₙ=y`. We prove it by induction on the length `n` of the sequence. Find an edge that
    splits the path into two parts of almost equal length: both `d x₀ x₁ + ... + d xₖ₋₁ xₖ` and
    `d xₖ₊₁ xₖ₊₂ + ... + d xₙ₋₁ xₙ` are less than or equal to `L / 2`.
    Then `d x₀ xₖ ≤ L`, `d xₖ xₖ₊₁ ≤ L`, and `d xₖ₊₁ xₙ ≤ L`, thus `d x₀ xₙ ≤ 2 * L`. -/
  rw [dist_ofPreNNDist, ← NNReal.coe_two, ← NNReal.coe_mul, NNReal.mul_iInf, NNReal.coe_le_coe]
  refine' le_ciInf fun l => _
  have hd₀_trans : Transitive fun x y => d x y = 0 := by
    intro a b c hab hbc
    rw [← nonpos_iff_eq_zero]
    simpa only [nonpos_iff_eq_zero, hab, hbc, dist_self c, max_self, mul_zero] using hd a b c c
  haveI : IsTrans X fun x y => d x y = 0 := ⟨hd₀_trans⟩
  induction' hn : length l using Nat.strong_induction_on with n ihn generalizing x y l
  simp only at ihn
  subst n
  set L := zipWith d (x::l) (l ++ [y])
  have hL_len : length L = length l + 1 := by simp
  rcases eq_or_ne (d x y) 0 with hd₀ | hd₀
  · simp only [hd₀, zero_le]
  rsuffices ⟨z, z', hxz, hzz', hz'y⟩ : ∃ z z' : X, d x z ≤ L.sum ∧ d z z' ≤ L.sum ∧ d z' y ≤ L.sum
  · exact (hd x z z' y).trans (mul_le_mul_left' (max_le hxz (max_le hzz' hz'y)) _)
  set s : Set ℕ := { m : ℕ | 2 * (take m L).sum ≤ L.sum }
  have hs₀ : 0 ∈ s := by simp
  have hsne : s.Nonempty := ⟨0, hs₀⟩
  obtain ⟨M, hMl, hMs⟩ : ∃ M ≤ length l, IsGreatest s M := by
    have hs_ub : length l ∈ upperBounds s := by
      intro m hm
      

Proof state:
X : Type u_1
d : X → X → ℝ≥0
dist_self : ∀ (x : X), d x x = 0
dist_comm : ∀ (x y : X), d x y = d y x
hd : ∀ (x₁ x₂ x₃ x₄ : X), d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))
hd₀_trans : Transitive fun x y => d x y = 0
this : IsTrans X fun x y => d x y = 0
x y : X
l : List X
ihn : ∀ m < length l, ∀ (x y : X) (l : List X), length l = m → d x y ≤ 2 * sum (zipWith d (x :: l) (l ++ [y]))
L : List ℝ≥0 := zipWith d (x :: l) (l ++ [y])
hL_len : length L = length l + 1
hd₀ : d x y ≠ 0
s : Set ℕ := {m | 2 * sum (take m L) ≤ sum L}
hs₀ : 0 ∈ s
hsne : Set.Nonempty s
m : ℕ
hm : m ∈ s
⊢ m ≤ length l

Next tactic:
rw [← not_lt, Nat.lt_iff_add_one_le, ← hL_len]
/-- Consider a function `d : X → X → ℝ≥0` such that `d x x = 0` and `d x y = d y x` for all `x`,
`y`. Let `dist` be the largest pseudometric distance such that `dist x y ≤ d x y`, see
`PseudoMetricSpace.ofPreNNDist`. Suppose that `d` satisfies the following triangle-like
inequality: `d x₁ x₄ ≤ 2 * max (d x₁ x₂, d x₂ x₃, d x₃ x₄)`. Then `d x y ≤ 2 * dist x y` for all
`x`, `y`. -/
theorem le_two_mul_dist_ofPreNNDist (d : X → X → ℝ≥0) (dist_self : ∀ x, d x x = 0)
    (dist_comm : ∀ x y, d x y = d y x)
    (hd : ∀ x₁ x₂ x₃ x₄, d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))) (x y : X) :
    ↑(d x y) ≤ 2 * @dist X
      (@PseudoMetricSpace.toDist X (PseudoMetricSpace.ofPreNNDist d dist_self dist_comm)) x y := by
  /- We need to show that `d x y` is at most twice the sum `L` of `d xᵢ xᵢ₊₁` over a path
    `x₀=x, ..., xₙ=y`. We prove it by induction on the length `n` of the sequence. Find an edge that
    splits the path into two parts of almost equal length: both `d x₀ x₁ + ... + d xₖ₋₁ xₖ` and
    `d xₖ₊₁ xₖ₊₂ + ... + d xₙ₋₁ xₙ` are less than or equal to `L / 2`.
    Then `d x₀ xₖ ≤ L`, `d xₖ xₖ₊₁ ≤ L`, and `d xₖ₊₁ xₙ ≤ L`, thus `d x₀ xₙ ≤ 2 * L`. -/
  rw [dist_ofPreNNDist, ← NNReal.coe_two, ← NNReal.coe_mul, NNReal.mul_iInf, NNReal.coe_le_coe]
  refine' le_ciInf fun l => _
  have hd₀_trans : Transitive fun x y => d x y = 0 := by
    intro a b c hab hbc
    rw [← nonpos_iff_eq_zero]
    simpa only [nonpos_iff_eq_zero, hab, hbc, dist_self c, max_self, mul_zero] using hd a b c c
  haveI : IsTrans X fun x y => d x y = 0 := ⟨hd₀_trans⟩
  induction' hn : length l using Nat.strong_induction_on with n ihn generalizing x y l
  simp only at ihn
  subst n
  set L := zipWith d (x::l) (l ++ [y])
  have hL_len : length L = length l + 1 := by simp
  rcases eq_or_ne (d x y) 0 with hd₀ | hd₀
  · simp only [hd₀, zero_le]
  rsuffices ⟨z, z', hxz, hzz', hz'y⟩ : ∃ z z' : X, d x z ≤ L.sum ∧ d z z' ≤ L.sum ∧ d z' y ≤ L.sum
  · exact (hd x z z' y).trans (mul_le_mul_left' (max_le hxz (max_le hzz' hz'y)) _)
  set s : Set ℕ := { m : ℕ | 2 * (take m L).sum ≤ L.sum }
  have hs₀ : 0 ∈ s := by simp
  have hsne : s.Nonempty := ⟨0, hs₀⟩
  obtain ⟨M, hMl, hMs⟩ : ∃ M ≤ length l, IsGreatest s M := by
    have hs_ub : length l ∈ upperBounds s := by
      intro m hm
      rw [← not_lt, Nat.lt_iff_add_one_le, ← hL_len]
      

Proof state:
X : Type u_1
d : X → X → ℝ≥0
dist_self : ∀ (x : X), d x x = 0
dist_comm : ∀ (x y : X), d x y = d y x
hd : ∀ (x₁ x₂ x₃ x₄ : X), d x₁ x₄ ≤ 2 * max (d x₁ x₂) (max (d x₂ x₃) (d x₃ x₄))
hd₀_trans : Transitive fun x y => d x y = 0
this : IsTrans X fun x y => d x y = 0
x y : X
l : List X
ihn : ∀ m < length l, ∀ (x y : X) (l : List X), length l = m → d x y ≤ 2 * sum (zipWith d (x :: l) (l ++ [y]))
L : List ℝ≥0 := zipWith d (x :: l) (l ++ [y])
hL_len : length L = length l + 1
hd₀ : d x y ≠ 0
s : Set ℕ := {m | 2 * sum (take m L) ≤ sum L}
hs₀ : 0 ∈ s
hsne : Set.Nonempty s
m : ℕ
hm : m ∈ s
⊢ ¬length L ≤ m

Next tactic:
intro hLm
/-- If `X` is a uniform space with countably generated uniformity filter, there exists a
`PseudoMetricSpace` structure compatible with the `UniformSpace` structure. Use
`UniformSpace.pseudoMetricSpace` or `UniformSpace.metricSpace` instead. -/
protected theorem UniformSpace.metrizable_uniformity (X : Type*) [UniformSpace X]
    [IsCountablyGenerated (𝓤 X)] : ∃ I : PseudoMetricSpace X, I.toUniformSpace = ‹_› := by
  classical
  /- Choose a fast decreasing antitone basis `U : ℕ → set (X × X)` of the uniformity filter `𝓤 X`.
    Define `d x y : ℝ≥0` to be `(1 / 2) ^ n`, where `n` is the minimal index of `U n` that
    separates `x` and `y`: `(x, y) ∉ U n`, or `0` if `x` is not separated from `y`. This function
    satisfies the assumptions of `PseudoMetricSpace.ofPreNNDist` and
    `PseudoMetricSpace.le_two_mul_dist_ofPreNNDist`, hence the distance given by the former pseudo
    metric space structure is Lipschitz equivalent to the `d`. Thus the uniformities generated by
    `d` and `dist` are equal. Since the former uniformity is equal to `𝓤 X`, the latter is equal to
    `𝓤 X` as well. -/
  

Proof state:
X✝ : Type u_1
X : Type u_2
inst✝¹ : UniformSpace X
inst✝ : IsCountablyGenerated (𝓤 X)
⊢ ∃ I, PseudoMetricSpace.toUniformSpace = inst✝¹

Next tactic:
obtain ⟨U, hU_symm, hU_comp, hB⟩ :
    ∃ U : ℕ → Set (X × X),
      (∀ n, SymmetricRel (U n)) ∧
        (∀ ⦃m n⦄, m < n → U n ○ (U n ○ U n) ⊆ U m) ∧ (𝓤 X).HasAntitoneBasis U := by
    rcases UniformSpace.has_seq_basis X with ⟨V, hB, hV_symm⟩
    rcases hB.subbasis_with_rel fun m =>
        hB.tendsto_smallSets.eventually
          (eventually_uniformity_iterate_comp_subset (hB.mem m) 2) with
      ⟨φ, -, hφ_comp, hφB⟩
    exact ⟨V ∘ φ, fun n => hV_symm _, hφ_comp, hφB⟩
/-- If `X` is a uniform space with countably generated uniformity filter, there exists a
`PseudoMetricSpace` structure compatible with the `UniformSpace` structure. Use
`UniformSpace.pseudoMetricSpace` or `UniformSpace.metricSpace` instead. -/
protected theorem UniformSpace.metrizable_uniformity (X : Type*) [UniformSpace X]
    [IsCountablyGenerated (𝓤 X)] : ∃ I : PseudoMetricSpace X, I.toUniformSpace = ‹_› := by
  classical
  /- Choose a fast decreasing antitone basis `U : ℕ → set (X × X)` of the uniformity filter `𝓤 X`.
    Define `d x y : ℝ≥0` to be `(1 / 2) ^ n`, where `n` is the minimal index of `U n` that
    separates `x` and `y`: `(x, y) ∉ U n`, or `0` if `x` is not separated from `y`. This function
    satisfies the assumptions of `PseudoMetricSpace.ofPreNNDist` and
    `PseudoMetricSpace.le_two_mul_dist_ofPreNNDist`, hence the distance given by the former pseudo
    metric space structure is Lipschitz equivalent to the `d`. Thus the uniformities generated by
    `d` and `dist` are equal. Since the former uniformity is equal to `𝓤 X`, the latter is equal to
    `𝓤 X` as well. -/
  obtain ⟨U, hU_symm, hU_comp, hB⟩ :
    ∃ U : ℕ → Set (X × X),
      (∀ n, SymmetricRel (U n)) ∧
        (∀ ⦃m n⦄, m < n → U n ○ (U n ○ U n) ⊆ U m) ∧ (𝓤 X).HasAntitoneBasis U := by
    

Proof state:
X✝ : Type u_1
X : Type u_2
inst✝¹ : UniformSpace X
inst✝ : IsCountablyGenerated (𝓤 X)
⊢ ∃ U, (∀ (n : ℕ), SymmetricRel (U n)) ∧ (∀ ⦃m n : ℕ⦄, m < n → U n ○ (U n ○ U n) ⊆ U m) ∧ HasAntitoneBasis (𝓤 X) U

Next tactic:
rcases UniformSpace.has_seq_basis X with ⟨V, hB, hV_symm⟩
/-- If `X` is a uniform space with countably generated uniformity filter, there exists a
`PseudoMetricSpace` structure compatible with the `UniformSpace` structure. Use
`UniformSpace.pseudoMetricSpace` or `UniformSpace.metricSpace` instead. -/
protected theorem UniformSpace.metrizable_uniformity (X : Type*) [UniformSpace X]
    [IsCountablyGenerated (𝓤 X)] : ∃ I : PseudoMetricSpace X, I.toUniformSpace = ‹_› := by
  classical
  /- Choose a fast decreasing antitone basis `U : ℕ → set (X × X)` of the uniformity filter `𝓤 X`.
    Define `d x y : ℝ≥0` to be `(1 / 2) ^ n`, where `n` is the minimal index of `U n` that
    separates `x` and `y`: `(x, y) ∉ U n`, or `0` if `x` is not separated from `y`. This function
    satisfies the assumptions of `PseudoMetricSpace.ofPreNNDist` and
    `PseudoMetricSpace.le_two_mul_dist_ofPreNNDist`, hence the distance given by the former pseudo
    metric space structure is Lipschitz equivalent to the `d`. Thus the uniformities generated by
    `d` and `dist` are equal. Since the former uniformity is equal to `𝓤 X`, the latter is equal to
    `𝓤 X` as well. -/
  obtain ⟨U, hU_symm, hU_comp, hB⟩ :
    ∃ U : ℕ → Set (X × X),
      (∀ n, SymmetricRel (U n)) ∧
        (∀ ⦃m n⦄, m < n → U n ○ (U n ○ U n) ⊆ U m) ∧ (𝓤 X).HasAntitoneBasis U := by
    rcases UniformSpace.has_seq_basis X with ⟨V, hB, hV_symm⟩
    

Proof state:
case intro.intro
X✝ : Type u_1
X : Type u_2
inst✝¹ : UniformSpace X
inst✝ : IsCountablyGenerated (𝓤 X)
V : ℕ → Set (X × X)
hB : HasAntitoneBasis (𝓤 X) V
hV_symm : ∀ (n : ℕ), SymmetricRel (V n)
⊢ ∃ U, (∀ (n : ℕ), SymmetricRel (U n)) ∧ (∀ ⦃m n : ℕ⦄, m < n → U n ○ (U n ○ U n) ⊆ U m) ∧ HasAntitoneBasis (𝓤 X) U

Next tactic:
rcases hB.subbasis_with_rel fun m =>
        hB.tendsto_smallSets.eventually
          (eventually_uniformity_iterate_comp_subset (hB.mem m) 2) with
      ⟨φ, -, hφ_comp, hφB⟩
/-- If `X` is a uniform space with countably generated uniformity filter, there exists a
`PseudoMetricSpace` structure compatible with the `UniformSpace` structure. Use
`UniformSpace.pseudoMetricSpace` or `UniformSpace.metricSpace` instead. -/
protected theorem UniformSpace.metrizable_uniformity (X : Type*) [UniformSpace X]
    [IsCountablyGenerated (𝓤 X)] : ∃ I : PseudoMetricSpace X, I.toUniformSpace = ‹_› := by
  classical
  /- Choose a fast decreasing antitone basis `U : ℕ → set (X × X)` of the uniformity filter `𝓤 X`.
    Define `d x y : ℝ≥0` to be `(1 / 2) ^ n`, where `n` is the minimal index of `U n` that
    separates `x` and `y`: `(x, y) ∉ U n`, or `0` if `x` is not separated from `y`. This function
    satisfies the assumptions of `PseudoMetricSpace.ofPreNNDist` and
    `PseudoMetricSpace.le_two_mul_dist_ofPreNNDist`, hence the distance given by the former pseudo
    metric space structure is Lipschitz equivalent to the `d`. Thus the uniformities generated by
    `d` and `dist` are equal. Since the former uniformity is equal to `𝓤 X`, the latter is equal to
    `𝓤 X` as well. -/
  obtain ⟨U, hU_symm, hU_comp, hB⟩ :
    ∃ U : ℕ → Set (X × X),
      (∀ n, SymmetricRel (U n)) ∧
        (∀ ⦃m n⦄, m < n → U n ○ (U n ○ U n) ⊆ U m) ∧ (𝓤 X).HasAntitoneBasis U := by
    rcases UniformSpace.has_seq_basis X with ⟨V, hB, hV_symm⟩
    rcases hB.subbasis_with_rel fun m =>
        hB.tendsto_smallSets.eventually
          (eventually_uniformity_iterate_comp_subset (hB.mem m) 2) with
      ⟨φ, -, hφ_comp, hφB⟩
    

Proof state:
case intro.intro.intro.intro.intro
X✝ : Type u_1
X : Type u_2
inst✝¹ : UniformSpace X
inst✝ : IsCountablyGenerated (𝓤 X)
V : ℕ → Set (X × X)
hB : HasAntitoneBasis (𝓤 X) V
hV_symm : ∀ (n : ℕ), SymmetricRel (V n)
φ : ℕ → ℕ
hφ_comp : ∀ ⦃m n : ℕ⦄, m < n → (fun x => V (φ n) ○ x)^[2] (V (φ n)) ⊆ V (φ m)
hφB : HasAntitoneBasis (𝓤 X) (V ∘ φ)
⊢ ∃ U, (∀ (n : ℕ), SymmetricRel (U n)) ∧ (∀ ⦃m n : ℕ⦄, m < n → U n ○ (U n ○ U n) ⊆ U m) ∧ HasAntitoneBasis (𝓤 X) U

Next tactic:
exact ⟨V ∘ φ, fun n => hV_symm _, hφ_comp, hφB⟩