l3lab/ntp-mathlib huggingface dataset:
/-- A reified version of the `Bertrand.main_inequality` below.
This is not best possible: it actually holds for 464 ≤ x.
-/
theorem real_main_inequality {x : ℝ} (n_large : (512 : ℝ) ≤ x) :
    x * (2 * x) ^ sqrt (2 * x) * 4 ^ (2 * x / 3) ≤ 4 ^ x := by
  let f : ℝ → ℝ := fun x => log x + sqrt (2 * x) * log (2 * x) - log 4 / 3 * x
  have hf' : ∀ x, 0 < x → 0 < x * (2 * x) ^ sqrt (2 * x) / 4 ^ (x / 3) := fun x h =>
    div_pos (mul_pos h (rpow_pos_of_pos (mul_pos two_pos h) _)) (rpow_pos_of_pos four_pos _)
  have hf : ∀ x, 0 < x → f x = log (x * (2 * x) ^ sqrt (2 * x) / 4 ^ (x / 3)) := by
    intro x h5
    have h6 := mul_pos (zero_lt_two' ℝ) h5
    have h7 := rpow_pos_of_pos h6 (sqrt (2 * x))
    rw [log_div (mul_pos h5 h7).ne' (rpow_pos_of_pos four_pos _).ne', log_mul h5.ne' h7.ne',
      log_rpow h6, log_rpow zero_lt_four, ← mul_div_right_comm, ← mul_div, mul_comm x]
  have h5 : 0 < x := lt_of_lt_of_le (by norm_num1) n_large
  rw [← div_le_one (rpow_pos_of_pos four_pos x), ← div_div_eq_mul_div, ← rpow_sub four_pos, ←
    mul_div 2 x, mul_div_left_comm, ← mul_one_sub, (by norm_num1 : (1 : ℝ) - 2 / 3 = 1 / 3),
    mul_one_div, ← log_nonpos_iff (hf' x h5), ← hf x h5]
  -- porting note: the proof was rewritten, because it was too slow
  have h : ConcaveOn ℝ (Set.Ioi 0.5) f := by
    apply ConcaveOn.sub
    apply ConcaveOn.add
    exact strictConcaveOn_log_Ioi.concaveOn.subset
      (Set.Ioi_subset_Ioi (by norm_num)) (convex_Ioi 0.5)
    convert ((strictConcaveOn_sqrt_mul_log_Ioi.concaveOn.comp_linearMap
      ((2 : ℝ) • LinearMap.id))) using 1
    · ext x
      simp only [Set.mem_Ioi, Set.mem_preimage, LinearMap.smul_apply,
        LinearMap.id_coe, id_eq, smul_eq_mul]
      rw [← mul_lt_mul_left (two_pos)]
      norm_num1
      rfl
    apply ConvexOn.smul
    refine div_nonneg (log_nonneg (by norm_num1)) (by norm_num1)
    exact convexOn_id (convex_Ioi (0.5 : ℝ))
  suffices ∃ x1 x2, 0.5 < x1 ∧ x1 < x2 ∧ x2 ≤ x ∧ 0 ≤ f x1 ∧ f x2 ≤ 0 by
    obtain ⟨x1, x2, h1, h2, h0, h3, h4⟩ := this
    exact (h.right_le_of_le_left'' h1 ((h1.trans h2).trans_le h0) h2 h0 (h4.trans h3)).trans h4
  refine' ⟨18, 512, by norm_num1, by norm_num1, n_large, _, _⟩
  · have : sqrt (2 * 18) = 6 := (sqrt_eq_iff_mul_self_eq_of_pos (by norm_num1)).mpr (by norm_num1)
    rw [hf, log_nonneg_iff, this]
    rw [one_le_div] <;> norm_num1
    apply le_trans _ (le_mul_of_one_le_left _ _) <;> norm_num1
    apply Real.rpow_le_rpow <;> norm_num1
    apply rpow_nonneg_of_nonneg; norm_num1
    apply rpow_pos_of_pos; norm_num1
    apply hf' 18; norm_num1
    norm_num1
  · have : sqrt (2 * 512) = 32 :=
      (sqrt_eq_iff_mul_self_eq_of_pos (by norm_num1)).mpr (by norm_num1)
    rw [hf, log_nonpos_iff (hf' _ _), this, div_le_one] <;> norm_num1
    · conv in 512 => equals 2 ^ 9 => norm_num1
      conv in 1024 => equals 2 ^ 10 => 

Proof state:
x : ℝ
n_large : 512 ≤ x
f : ℝ → ℝ := fun x => log x + sqrt (2 * x) * log (2 * x) - log 4 / 3 * x
hf' : ∀ (x : ℝ), 0 < x → 0 < x * (2 * x) ^ sqrt (2 * x) / 4 ^ (x / 3)
hf : ∀ (x : ℝ), 0 < x → f x = log (x * (2 * x) ^ sqrt (2 * x) / 4 ^ (x / 3))
h5 : 0 < x
h : ConcaveOn ℝ (Set.Ioi 0.5) f
this : sqrt (2 * 512) = 32
⊢ 1024 = 2 ^ 10

Next tactic:
norm_num1
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  

Proof state:
n : ℕ
n_large : 512 ≤ n
⊢ n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n

Next tactic:
rw [← @cast_le ℝ]
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  

Proof state:
n : ℕ
n_large : 512 ≤ n
⊢ ↑(n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3)) ≤ ↑(4 ^ n)

Next tactic:
simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  

Proof state:
n : ℕ
n_large : 512 ≤ n
⊢ ↑n * (↑2 * ↑n) ^ ↑(sqrt (2 * n)) * ↑4 ^ ↑(2 * n / 3) ≤ ↑4 ^ ↑n

Next tactic:
refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by 

Proof state:
n : ℕ
n_large : 512 ≤ n
⊢ 512 ≤ ↑n

Next tactic:
exact_mod_cast n_large
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
  

Proof state:
n : ℕ
n_large : 512 ≤ n
⊢ ↑n * (↑2 * ↑n) ^ ↑(sqrt (2 * n)) * ↑4 ^ ↑(2 * n / 3) ≤ ↑n * (2 * ↑n) ^ Real.sqrt (2 * ↑n) * 4 ^ (2 * ↑n / 3)

Next tactic:
gcongr
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
  gcongr
  · 

Proof state:
case h₁.h.hx
n : ℕ
n_large : 512 ≤ n
⊢ 1 ≤ ↑2 * ↑n

Next tactic:
have n2_pos : 0 < 2 * n := by positivity
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
  gcongr
  · have n2_pos : 0 < 2 * n := by 

Proof state:
n : ℕ
n_large : 512 ≤ n
⊢ 0 < 2 * n

Next tactic:
positivity
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
  gcongr
  · have n2_pos : 0 < 2 * n := by positivity
    

Proof state:
case h₁.h.hx
n : ℕ
n_large : 512 ≤ n
n2_pos : 0 < 2 * n
⊢ 1 ≤ ↑2 * ↑n

Next tactic:
exact mod_cast n2_pos
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
  gcongr
  · have n2_pos : 0 < 2 * n := by positivity
    exact mod_cast n2_pos
  · 

Proof state:
case h₁.h.hyz
n : ℕ
n_large : 512 ≤ n
⊢ ↑(sqrt (2 * n)) ≤ Real.sqrt (2 * ↑n)

Next tactic:
exact_mod_cast Real.nat_sqrt_le_real_sqrt
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
  gcongr
  · have n2_pos : 0 < 2 * n := by positivity
    exact mod_cast n2_pos
  · exact_mod_cast Real.nat_sqrt_le_real_sqrt
  · 

Proof state:
case h₂.hx
n : ℕ
n_large : 512 ≤ n
⊢ 1 ≤ ↑4

Next tactic:
norm_num1
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
  gcongr
  · have n2_pos : 0 < 2 * n := by positivity
    exact mod_cast n2_pos
  · exact_mod_cast Real.nat_sqrt_le_real_sqrt
  · norm_num1
  · 

Proof state:
case h₂.hyz
n : ℕ
n_large : 512 ≤ n
⊢ ↑(2 * n / 3) ≤ 2 * ↑n / 3

Next tactic:
exact cast_div_le.trans (by norm_cast)
/-- The inequality which contradicts Bertrand's postulate, for large enough `n`.
-/
theorem bertrand_main_inequality {n : ℕ} (n_large : 512 ≤ n) :
    n * (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) ≤ 4 ^ n := by
  rw [← @cast_le ℝ]
  simp only [cast_add, cast_one, cast_mul, cast_pow, ← Real.rpow_nat_cast]
  refine' _root_.trans ?_ (Bertrand.real_main_inequality (by exact_mod_cast n_large))
  gcongr
  · have n2_pos : 0 < 2 * n := by positivity
    exact mod_cast n2_pos
  · exact_mod_cast Real.nat_sqrt_le_real_sqrt
  · norm_num1
  · exact cast_div_le.trans (by 

Proof state:
n : ℕ
n_large : 512 ≤ n
⊢ ↑(2 * n) / ↑3 ≤ 2 * ↑n / 3

Next tactic:
norm_cast
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  

Proof state:
n : ℕ
n_large : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
⊢ centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (Nat.factorization (centralBinom n)) p

Next tactic:
refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  

Proof state:
n : ℕ
n_large : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
⊢ ∏ p in Finset.range (2 * n / 3 + 1), p ^ (Nat.factorization (centralBinom n)) p =
    ∏ p in Finset.range (2 * n + 1), p ^ (Nat.factorization (centralBinom n)) p

Next tactic:
apply Finset.prod_subset
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · 

Proof state:
case h
n : ℕ
n_large : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
⊢ Finset.range (2 * n / 3 + 1) ⊆ Finset.range (2 * n + 1)

Next tactic:
exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
  

Proof state:
case hf
n : ℕ
n_large : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
⊢ ∀ x ∈ Finset.range (2 * n + 1), x ∉ Finset.range (2 * n / 3 + 1) → x ^ (Nat.factorization (centralBinom n)) x = 1

Next tactic:
intro x hx h2x
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
  intro x hx h2x
  

Proof state:
case hf
n : ℕ
n_large : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
x : ℕ
hx : x ∈ Finset.range (2 * n + 1)
h2x : x ∉ Finset.range (2 * n / 3 + 1)
⊢ x ^ (Nat.factorization (centralBinom n)) x = 1

Next tactic:
rw [Finset.mem_range, lt_succ_iff] at hx h2x
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
  intro x hx h2x
  rw [Finset.mem_range, lt_succ_iff] at hx h2x
  

Proof state:
case hf
n : ℕ
n_large : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
x : ℕ
hx : x ≤ 2 * n
h2x : ¬x ≤ 2 * n / 3
⊢ x ^ (Nat.factorization (centralBinom n)) x = 1

Next tactic:
rw [not_le, div_lt_iff_lt_mul' three_pos, mul_comm x] at h2x
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
  intro x hx h2x
  rw [Finset.mem_range, lt_succ_iff] at hx h2x
  rw [not_le, div_lt_iff_lt_mul' three_pos, mul_comm x] at h2x
  

Proof state:
case hf
n : ℕ
n_large : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
x : ℕ
hx : x ≤ 2 * n
h2x : 2 * n < 3 * x
⊢ x ^ (Nat.factorization (centralBinom n)) x = 1

Next tactic:
replace no_prime := not_exists.mp no_prime x
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
  intro x hx h2x
  rw [Finset.mem_range, lt_succ_iff] at hx h2x
  rw [not_le, div_lt_iff_lt_mul' three_pos, mul_comm x] at h2x
  replace no_prime := not_exists.mp no_prime x
  

Proof state:
case hf
n : ℕ
n_large : 2 < n
x : ℕ
hx : x ≤ 2 * n
h2x : 2 * n < 3 * x
no_prime : ¬(Nat.Prime x ∧ n < x ∧ x ≤ 2 * n)
⊢ x ^ (Nat.factorization (centralBinom n)) x = 1

Next tactic:
rw [← and_assoc, not_and', not_and_or, not_lt] at no_prime
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
  intro x hx h2x
  rw [Finset.mem_range, lt_succ_iff] at hx h2x
  rw [not_le, div_lt_iff_lt_mul' three_pos, mul_comm x] at h2x
  replace no_prime := not_exists.mp no_prime x
  rw [← and_assoc, not_and', not_and_or, not_lt] at no_prime
  

Proof state:
case hf
n : ℕ
n_large : 2 < n
x : ℕ
hx : x ≤ 2 * n
h2x : 2 * n < 3 * x
no_prime : x ≤ 2 * n → ¬Nat.Prime x ∨ x ≤ n
⊢ x ^ (Nat.factorization (centralBinom n)) x = 1

Next tactic:
cases' no_prime hx with h h
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
  intro x hx h2x
  rw [Finset.mem_range, lt_succ_iff] at hx h2x
  rw [not_le, div_lt_iff_lt_mul' three_pos, mul_comm x] at h2x
  replace no_prime := not_exists.mp no_prime x
  rw [← and_assoc, not_and', not_and_or, not_lt] at no_prime
  cases' no_prime hx with h h
  · 

Proof state:
case hf.inl
n : ℕ
n_large : 2 < n
x : ℕ
hx : x ≤ 2 * n
h2x : 2 * n < 3 * x
no_prime : x ≤ 2 * n → ¬Nat.Prime x ∨ x ≤ n
h : ¬Nat.Prime x
⊢ x ^ (Nat.factorization (centralBinom n)) x = 1

Next tactic:
rw [factorization_eq_zero_of_non_prime n.centralBinom h, Nat.pow_zero]
/-- A lemma that tells us that, in the case where Bertrand's postulate does not hold, the prime
factorization of the central binomial coefficent only has factors at most `2 * n / 3 + 1`.
-/
theorem centralBinom_factorization_small (n : ℕ) (n_large : 2 < n)
    (no_prime : ¬∃ p : ℕ, p.Prime ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n = ∏ p in Finset.range (2 * n / 3 + 1), p ^ (centralBinom n).factorization p := by
  refine' (Eq.trans _ n.prod_pow_factorization_centralBinom).symm
  apply Finset.prod_subset
  · exact Finset.range_subset.2 (add_le_add_right (Nat.div_le_self _ _) _)
  intro x hx h2x
  rw [Finset.mem_range, lt_succ_iff] at hx h2x
  rw [not_le, div_lt_iff_lt_mul' three_pos, mul_comm x] at h2x
  replace no_prime := not_exists.mp no_prime x
  rw [← and_assoc, not_and', not_and_or, not_lt] at no_prime
  cases' no_prime hx with h h
  · rw [factorization_eq_zero_of_non_prime n.centralBinom h, Nat.pow_zero]
  · 

Proof state:
case hf.inr
n : ℕ
n_large : 2 < n
x : ℕ
hx : x ≤ 2 * n
h2x : 2 * n < 3 * x
no_prime : x ≤ 2 * n → ¬Nat.Prime x ∨ x ≤ n
h : x ≤ n
⊢ x ^ (Nat.factorization (centralBinom n)) x = 1

Next tactic:
rw [factorization_centralBinom_of_two_mul_self_lt_three_mul n_large h h2x, Nat.pow_zero]
/-- An upper bound on the central binomial coefficient used in the proof of Bertrand's postulate.
The bound splits the prime factors of `centralBinom n` into those
1. At most `sqrt (2 * n)`, which contribute at most `2 * n` for each such prime.
2. Between `sqrt (2 * n)` and `2 * n / 3`, which contribute at most `4^(2 * n / 3)` in total.
3. Between `2 * n / 3` and `n`, which do not exist.
4. Between `n` and `2 * n`, which would not exist in the case where Bertrand's postulate is false.
5. Above `2 * n`, which do not exist.
-/
theorem centralBinom_le_of_no_bertrand_prime (n : ℕ) (n_big : 2 < n)
    (no_prime : ¬∃ p : ℕ, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n ≤ (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) := by
  

Proof state:
n : ℕ
n_big : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
⊢ centralBinom n ≤ (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3)

Next tactic:
have n_pos : 0 < n := (Nat.zero_le _).trans_lt n_big
/-- An upper bound on the central binomial coefficient used in the proof of Bertrand's postulate.
The bound splits the prime factors of `centralBinom n` into those
1. At most `sqrt (2 * n)`, which contribute at most `2 * n` for each such prime.
2. Between `sqrt (2 * n)` and `2 * n / 3`, which contribute at most `4^(2 * n / 3)` in total.
3. Between `2 * n / 3` and `n`, which do not exist.
4. Between `n` and `2 * n`, which would not exist in the case where Bertrand's postulate is false.
5. Above `2 * n`, which do not exist.
-/
theorem centralBinom_le_of_no_bertrand_prime (n : ℕ) (n_big : 2 < n)
    (no_prime : ¬∃ p : ℕ, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n ≤ (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) := by
  have n_pos : 0 < n := (Nat.zero_le _).trans_lt n_big
  

Proof state:
n : ℕ
n_big : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
n_pos : 0 < n
⊢ centralBinom n ≤ (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3)

Next tactic:
have n2_pos : 1 ≤ 2 * n := mul_pos (zero_lt_two' ℕ) n_pos
/-- An upper bound on the central binomial coefficient used in the proof of Bertrand's postulate.
The bound splits the prime factors of `centralBinom n` into those
1. At most `sqrt (2 * n)`, which contribute at most `2 * n` for each such prime.
2. Between `sqrt (2 * n)` and `2 * n / 3`, which contribute at most `4^(2 * n / 3)` in total.
3. Between `2 * n / 3` and `n`, which do not exist.
4. Between `n` and `2 * n`, which would not exist in the case where Bertrand's postulate is false.
5. Above `2 * n`, which do not exist.
-/
theorem centralBinom_le_of_no_bertrand_prime (n : ℕ) (n_big : 2 < n)
    (no_prime : ¬∃ p : ℕ, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n ≤ (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) := by
  have n_pos : 0 < n := (Nat.zero_le _).trans_lt n_big
  have n2_pos : 1 ≤ 2 * n := mul_pos (zero_lt_two' ℕ) n_pos
  

Proof state:
n : ℕ
n_big : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
n_pos : 0 < n
n2_pos : 1 ≤ 2 * n
⊢ centralBinom n ≤ (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3)

Next tactic:
let S := (Finset.range (2 * n / 3 + 1)).filter Nat.Prime
/-- An upper bound on the central binomial coefficient used in the proof of Bertrand's postulate.
The bound splits the prime factors of `centralBinom n` into those
1. At most `sqrt (2 * n)`, which contribute at most `2 * n` for each such prime.
2. Between `sqrt (2 * n)` and `2 * n / 3`, which contribute at most `4^(2 * n / 3)` in total.
3. Between `2 * n / 3` and `n`, which do not exist.
4. Between `n` and `2 * n`, which would not exist in the case where Bertrand's postulate is false.
5. Above `2 * n`, which do not exist.
-/
theorem centralBinom_le_of_no_bertrand_prime (n : ℕ) (n_big : 2 < n)
    (no_prime : ¬∃ p : ℕ, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n) :
    centralBinom n ≤ (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3) := by
  have n_pos : 0 < n := (Nat.zero_le _).trans_lt n_big
  have n2_pos : 1 ≤ 2 * n := mul_pos (zero_lt_two' ℕ) n_pos
  let S := (Finset.range (2 * n / 3 + 1)).filter Nat.Prime
  

Proof state:
n : ℕ
n_big : 2 < n
no_prime : ¬∃ p, Nat.Prime p ∧ n < p ∧ p ≤ 2 * n
n_pos : 0 < n
n2_pos : 1 ≤ 2 * n
S : Finset ℕ := Finset.filter Nat.Prime (Finset.range (2 * n / 3 + 1))
⊢ centralBinom n ≤ (2 * n) ^ sqrt (2 * n) * 4 ^ (2 * n / 3)

Next tactic:
let f x := x ^ n.centralBinom.factorization x