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Proof step 27 from proof bertrand-prime-factors (previous step)

all_prime_divisors(binom_coef(2 * i, i)) = ints0 + ints2 + ints3 + ints, 0 < i, 1 < i, ordered≤(ints0 + ints2 + ints3 + ints), j2 ∈ ints, ∀ i0. i0 ∈ ints2 ∨ i0 ∈ ints3 ∨ i0 ∈ ints → sqrt(2 * i) < i0, ∀ i0. i0 ∈ ints3 ∨ i0 ∈ ints → (2 * i) / 3 < i0, ∀ i1. i1 ∈ ints0 → i1 < sqrt(2 * i) ∨ i1 = sqrt(2 * i), ∀ i1. i1 ∈ ints2 → i1 < (2 * i) / 3 ∨ i1 = (2 * i) / 3, ∀ i1. i1 ∈ ints → i < i1, ∀ i1. i1 ∈ ints3 → i1 < i ∨ i1 = i ⊦ i < j2 ∧ j2 ≤ 2 * i

User applied rule all left

And obtained the following new goals

Proof step 28:
all_prime_divisors(binom_coef(2 * i, i)) = ints0 + ints2 + ints3 + ints, 0 < i, 1 < i, ordered≤(ints0 + ints2 + ints3 + ints), j2 ∈ ints, ∀ i0. i0 ∈ ints2 ∨ i0 ∈ ints3 ∨ i0 ∈ ints → sqrt(2 * i) < i0, ∀ i0. i0 ∈ ints3 ∨ i0 ∈ ints → (2 * i) / 3 < i0, ∀ i1. i1 ∈ ints0 → i1 < sqrt(2 * i) ∨ i1 = sqrt(2 * i), ∀ i1. i1 ∈ ints2 → i1 < (2 * i) / 3 ∨ i1 = (2 * i) / 3, j2 ∈ ints → i < j2, ∀ i1. i1 ∈ ints → i < i1, ∀ i1. i1 ∈ ints3 → i1 < i ∨ i1 = i ⊦ i < j2 ∧ j2 ≤ 2 * i

Rule argument

pos: left 10 terms: j2 abort: F discard: F computed: T

Additional information

Used terms: j2