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Proof step 4 from proof max_prime_divisor-ge2 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 6:
¬ prime(i) ∧ 2 ≤ i → (∃ i0. max_prime_divisor(i) * i0 = i), 2 ≤ i, i / max_prime_divisor(i) < 2, 0 ≤ i, 0 < i, i / max_prime_divisor(i) < i ⊦ prime(i)

Rule argument

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

Additional information

Used terms: i, max_prime_divisor(i)