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Proof step 6 from proof BertrandsPostulate (previous step)

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 insert lemma

And obtained the following new goals

Proof step 7:
all_prime_divisors(binom_coef(2 * i, i)) = ints + ints0 + ints1 + ints2 ∧ (∀ j. j ∈ ints → j ≤ sqrt(2 * i)) ∧ (∀ j. j ∈ ints0 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3) ∧ (∀ j. j ∈ ints1 → (2 * i) / 3 < j ∧ j ≤ i) ∧ (∀ j. j ∈ ints2 → i < j ∧ j ≤ 2 * i), 2 < i ⊦ binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3) * Π ints2
Proof step 8:
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

Rule argument

spec: inst: name: bertrand-middle-coef-smaller seq: all_prime_divisors(binom_coef(2 * i, i)) = ints + ints0 + ints1 + ints2 ∧ (∀ j. j ∈ ints → j ≤ sqrt(2 * i)) ∧ (∀ j. j ∈ ints0 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3) ∧ (∀ j. j ∈ ints1 → (2 * i) / 3 < j ∧ j ≤ i) ∧ (∀ j. j ∈ ints2 → i < j ∧ j ≤ 2 * i), 2 < i ⊦ binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3) * Π ints2 subst: [i, ints, ints0, ints1, ints2] → [i, ints2, ints1, ints0, ints] flags: current, []

Additional information

Used terms: i, ints2, ints1, ints0, ints