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Proof step 117 from proof all_prime_divisors-div (previous step)

all_prime_divisors(((i * i0) / j0) / i) = all_prime_divisors((i * i0) / j0) -1l i, 0 ≤ i0, 2 ≤ i * i0, 2 ≤ i0, prime(j2), j2 | i0, j2 < i0, j0 < i * i0, j0 | i * i0, prime(j0), i | (i * i0) / j0, prime(i), 0 < i, 0 < i0, ∀ j. j2 < j ∧ j < i0 ∧ prime(j) → ¬ j | i0, ∀ j1. j0 < j1 ∧ j1 < i * i0 ∧ prime(j1) → ¬ j1 | i * i0, ∀ j0, i1. i→n(j0) < i→n(i * i0) → i1 | j0 ∧ prime(i1) ∧ 0 < j0 → all_prime_divisors(j0 / i1) = all_prime_divisors(j0) -1l i1 ⊦ all_prime_divisors(i0 / j2) + j2 ' = (all_prime_divisors((i * i0) / j0) -1l i) + j0 ', j2 = j0, prime(i0), prime(i * i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i * i0, j0] flags: rewrite

Additional information

Used terms: i * i0, j0