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Proof step 3 from proof max_prime_divisor-divides-02 (previous step)

0 < i, prime(j), j | i / max_prime_divisor(i), 2 ≤ i, 0 ≤ i ⊦ prime(i), j | i

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) | i
Proof step 5:
0 < i, prime(j), j | i / max_prime_divisor(i), 0 ≤ i, 2 ≤ i, ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) | i ⊦ prime(i), j | i

Rule argument

spec: <toplevel> name: max_prime_divisor-divides seq: ⊦ ¬ prime(i) ∧ 2 ≤ i → max_prime_divisor(i) | i subst: [i] → [i]

Additional information

Used terms: i