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Proof step 29 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, i < j2, 2 * i < j2, 0 < j2, 1 < j2, ∀ i1. i1 ∈ ints → i < i1, ∀ i1. i1 ∈ ints3 → i1 < i ∨ i1 = i, ∀ 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 ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: j2 ∈ all_prime_divisors(binom_coef(2 * i, i))

Additional information

Used simplifier rules: middle-coef-divides-not: ⊦ 2 * i < j ∧ prime(j) ∧ 0 < i → ¬ j | binom_coef(2 * i, i);%in-12: ⊦ i ∈ ints → i ∈ ints0 + ints; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%all_prime_divisors-in-01: ⊦ j ∈ all_prime_divisors(i) ↔ prime(j) ∧ j | i ∧ 2 ≤ i; from specification intlist%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used formulas: j2 ∈ all_prime_divisors(binom_coef(2 * i, i))