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Proof step 16 from proof bertrand-prime-factors (previous step)

all_prime_divisors(binom_coef(2 * i, i)) = ints0 + ints2 + ints3 + ints, ordered≤(ints0 + ints2 + ints3 + ints), 1 < i, 0 < i, ∀ i1. i1 ∈ ints3 → i1 < i ∨ i1 = i, ∀ i1. i1 ∈ ints → i < i1, ∀ i1. i1 ∈ ints2 → i1 < (2 * i) / 3 ∨ i1 = (2 * i) / 3, ∀ i1. i1 ∈ ints0 → i1 < sqrt(2 * i) ∨ i1 = sqrt(2 * i), ∀ i0. i0 ∈ ints3 ∨ i0 ∈ ints → (2 * i) / 3 < i0, ∀ i0. i0 ∈ ints2 ∨ i0 ∈ ints3 ∨ i0 ∈ ints → sqrt(2 * i) < i0 ⊦ j ∈ ints0 → j ≤ sqrt(2 * i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
all_prime_divisors(binom_coef(2 * i, i)) = ints0 + ints2 + ints3 + ints, j ∈ ints0, ordered≤(ints0 + ints2 + ints3 + ints), 1 < i, 0 < i, sqrt(2 * i) < j, ∀ i1. i1 ∈ ints3 → i1 < i ∨ i1 = i, ∀ i1. i1 ∈ ints → i < i1, ∀ i1. i1 ∈ ints2 → i1 < (2 * i) / 3 ∨ i1 = (2 * i) / 3, ∀ i1. i1 ∈ ints0 → i1 < sqrt(2 * i) ∨ i1 = sqrt(2 * i), ∀ i0. i0 ∈ ints3 ∨ i0 ∈ ints → (2 * i) / 3 < i0, ∀ i0. i0 ∈ ints2 ∨ i0 ∈ ints3 ∨ i0 ∈ ints → sqrt(2 * i) < i0 ⊦

Rule argument

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs