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Proof step 8 from proof BertrandsPostulate (previous step)

binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3) * Π ints, all_prime_divisors(binom_coef(2 * i, i)) = ints2 + ints1 + ints0 + ints, 0 < i, 2 < i, ∀ j. j ∈ ints → i < j ∧ j ≤ 2 * i, ∀ j. j ∈ ints0 → (2 * i) / 3 < j ∧ j ≤ i, ∀ j. j ∈ ints1 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3, ∀ j. j ∈ ints2 → j ≤ sqrt(2 * i) ⊦ ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i

User applied rule cut formula

And obtained the following new goals

Proof step 9:
ints = [], binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3) * Π ints, all_prime_divisors(binom_coef(2 * i, i)) = ints2 + ints1 + ints0 + ints, 0 < i, 2 < i, ∀ j. j ∈ ints → i < j ∧ j ≤ 2 * i, ∀ j. j ∈ ints0 → (2 * i) / 3 < j ∧ j ≤ i, ∀ j. j ∈ ints1 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3, ∀ j. j ∈ ints2 → j ≤ sqrt(2 * i) ⊦ ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i
Proof step 25:
binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3) * Π ints, all_prime_divisors(binom_coef(2 * i, i)) = ints2 + ints1 + ints0 + ints, 0 < i, 2 < i, ∀ j. j ∈ ints → i < j ∧ j ≤ 2 * i, ∀ j. j ∈ ints0 → (2 * i) / 3 < j ∧ j ≤ i, ∀ j. j ∈ ints1 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3, ∀ j. j ∈ ints2 → j ≤ sqrt(2 * i) ⊦ ints = [], ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i

Rule argument

fma: ints = []

Additional information

Used formulas: ints = []