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Proof step 13 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 ⊦ ∃ 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)

User applied rule exists right

And obtained the following new goals

Proof step 14:
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 ⊦ all_prime_divisors(binom_coef(2 * i, i)) = ints0 + ints2 + ints3 + ints ∧ (∀ j. j ∈ ints0 → j ≤ sqrt(2 * i)) ∧ (∀ j. j ∈ ints2 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3) ∧ (∀ j. j ∈ ints3 → (2 * i) / 3 < j ∧ j ≤ i) ∧ (∀ j. j ∈ ints → i < j ∧ j ≤ 2 * i)

Rule argument

pos: right 1 terms: ints0, ints2, ints3, ints abort: F discard: T computed: T

Additional information

Used terms: ints0, ints2, ints3, ints