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Proof step 9 from proof bertrand-prime-factors (previous step)

∃ ints0, ints1. ints = ints0 + ints1 ∧ (∀ i1. i1 ∈ ints0 → i1 < (2 * i) / 3 ∨ i1 = (2 * i) / 3) ∧ (∀ i1. i1 ∈ ints1 → (2 * i) / 3 < i1), all_prime_divisors(binom_coef(2 * i, i)) = ints0 + ints, ordered≤(ints0 + ints), 1 < i, 0 < i, ∀ i1. i1 ∈ ints0 → i1 < sqrt(2 * i) ∨ i1 = sqrt(2 * i), ∀ i1. i1 ∈ ints → sqrt(2 * i) < i1 ⊦ ∃ ints, ints0, ints1, ints2. 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
all_prime_divisors(binom_coef(2 * i, i)) = ints0 + ints2 + ints1, ordered≤(ints0 + ints2 + ints1), 1 < i, 0 < i, ∀ i1. i1 ∈ ints2 → i1 < (2 * i) / 3 ∨ i1 = (2 * i) / 3, ∀ i1. i1 ∈ ints1 → (2 * i) / 3 < i1, ∀ i1. i1 ∈ ints0 → i1 < sqrt(2 * i) ∨ i1 = sqrt(2 * i), ∀ i0. i0 ∈ ints2 ∨ i0 ∈ ints1 → sqrt(2 * i) < i0 ⊦ ∃ ints, ints0, ints1, ints2. 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)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%in-append: ⊦ i ∈ ints + ints0 ↔ ¬ ¬(i ∈ ints ∨ i ∈ ints0); from specification list[intlista]