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Proof step 5 from proof max_prime_divisor-divides-03 (previous step)

0 < i, prime(j), j | i, 0 ≤ i, 2 ≤ i, ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) | i ⊦ j = max_prime_divisor(i), prime(i), j | i / max_prime_divisor(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 7:
0 < i, prime(j), j | i, 2 ≤ i, 0 ≤ i, ¬ prime(i) ∧ 2 ≤ i → (∃ i0. max_prime_divisor(i) * i0 = i) ⊦ j = max_prime_divisor(i), prime(i), j | i / max_prime_divisor(i)

Rule argument

spec: <toplevel> name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i, max_prime_divisor(i)] flags: <all paths>

Additional information

Used terms: i, max_prime_divisor(i)