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Proof step 3 from proof max_prime_divisor-nls-zero (previous step)

max_prime_divisor(i) < 0, 0 < i, 2 ≤ i, 0 ≤ i ⊦ prime(i)

User applied rule apply lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used terms: i