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Proof step 8 from proof divides-prime (previous step)

0 < j, i | j, 2 ≤ j, 0 ≤ j, ∀ i. 1 < i ∧ i < j → ¬ i | j ⊦ abs(i) = 1, abs(i) = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 10:
0 < j, 0 ≤ j, 2 ≤ j, ∀ i. 1 < i ∧ i < j → ¬ i | j, ∃ i0. i * i0 = j ⊦ abs(i) = j, abs(i) = 1

Rule argument

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

Additional information

Used terms: j, i