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Proof step 9 from proof max_prime_divisor-nls-03 (previous step)

j < i, j < i / j, 0 < i, prime(j), prime(i / j), j | i, 2 ≤ i, 0 ≤ 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 10:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 11:
0 < i, j < i / j, j < i, prime(i / j), prime(j), 0 ≤ i, 2 ≤ i, ∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i, ∃ i0. j * i0 = i ⊦ prime(i)

Rule argument

spec: <toplevel> name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i, j] flags: <all paths>

Additional information

Used terms: i, j