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Proof step 15 from proof BertrandsPostulate (previous step)

4 ^ ((2 * i) / 3) * (2 * i) ^ (sqrt(2 * i) - 1) * 2 * i = 4 ^ ((2 * i) / 3) * (2 * i) ^ sqrt(2 * i), 4 ^ i ≤ 4 ^ ((2 * i) / 3) * (2 * i) ^ (sqrt(2 * i) - 1) * 2 * i, all_prime_divisors(binom_coef(2 * i, i)) = ints2 + ints1 + ints0, 0 < i, 2 < i, binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3), 4 ^ i ≤ binom_coef(2 * i, i) * 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 equation

And obtained the following new goals

Proof step 16:
4 ^ i ≤ 4 ^ ((2 * i) / 3) * (2 * i) ^ sqrt(2 * i), all_prime_divisors(binom_coef(2 * i, i)) = ints2 + ints1 + ints0, 0 < i, 2 < i, binom_coef(2 * i, i) ≤ (2 * i) ^ (sqrt(2 * i) - 1) * 4 ^ ((2 * i) / 3), 4 ^ i ≤ binom_coef(2 * i, i) * 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

pos: left 1 flags: discard

Proof step 15 from proof BertrandsPostulate (previous step)

User applied rule insert equation

And obtained the following new goals

Rule argument

Additional information