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Proof step 13 from proof all_prime_divisors-div (previous step)

2 ≤ j, 0 < j, prime(i), i | j, 0 ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → i | j0 ∧ prime(i) ∧ 0 < j0 → all_prime_divisors(j0 / i) = all_prime_divisors(j0) -1l i ⊦ all_prime_divisors(j / i) = (all_prime_divisors(j / max_prime_divisor(j)) + max_prime_divisor(j) ') -1l i, prime(j)

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [j0, i] → [j / max_prime_divisor(j), i]

Additional information

Used simplifier rules: labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%max_prime_divisor-not-zero: ⊦ ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) ≠ 0; from specification int-gcd%abs: j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); from specification int-div%max_prime_divisor-abs-01: ⊦ ¬ prime(i) ∧ 2 ≤ i → abs(max_prime_divisor(i)) = max_prime_divisor(i); from specification int-gcd%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%max_prime_divisor-nle-01: ⊦ ¬ prime(i) ∧ 2 ≤ i → ¬ i ≤ i / max_prime_divisor(i); from specification int-gcd Used terms: j / max_prime_divisor(j), i