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Proof step 26 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) ⊦ ints = [], prime(ints.first) ∧ i < ints.first ∧ ints.first ≤ 2 * i

User applied rule insert elim lemma

And obtained the following new goals

Proof step 27:
ints = i0 + ints3, 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 = [], prime(i0) ∧ i < i0 ∧ i0 ≤ 2 * i

Rule argument

spec: list-data inst: intlista name: elim-1 seq: ⊦ ints ≠ [] → (i = ints.first ∧ ints0 = ints.rest ↔ ints = i + ints0) subst: [ints, i, ints0] → [ints, i0, ints3] flags: []

Additional information

Used simplifier rules: ⊦ ints ≠ [] → (i = ints.first ∧ ints0 = ints.rest ↔ ints = i + ints0); (elim) Used terms: ints, i0, ints3