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Proof step 8 from proof max_prime_divisor-nls-03 (previous step)

j < i / j, prime(i / j), 2 ≤ i, 0 ≤ i, 0 < i, prime(j), j | i, j < i, ∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i ⊦ prime(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 10:
j < i / j, prime(i / j), 2 ≤ i, 0 ≤ i, 0 < i, prime(j), ∃ i0. j * i0 = i, j < i, ∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i ⊦ prime(i)

Rule argument

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

Additional information

Used terms: i, j