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

0 < i, 2 < i / max_prime_divisor(i), i < i / max_prime_divisor(i), 0 < i / max_prime_divisor(i), 1 < i / max_prime_divisor(i), 0 ≤ i, 2 ≤ i ⊦ prime(i)

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ ¬ prime(i) ∧ 2 ≤ i → (∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i))
Proof step 5:
1 < i / max_prime_divisor(i), 0 < i / max_prime_divisor(i), i < i / max_prime_divisor(i), 2 < i / max_prime_divisor(i), 0 < i, 2 ≤ i, 0 ≤ i, ¬ prime(i) ∧ 2 ≤ i → (∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i)) ⊦ prime(i)

Rule argument

spec: <toplevel> name: prime-ex-max seq: ⊦ ¬ prime(i) ∧ 2 ≤ i → (∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i)) subst: [i] → [i]

Additional information

Used terms: i