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Proof step 16 from proof prime-def-mod (previous step)

0 < j, i < j, 0 < i, 1 < i, i | j, 2 ≤ j, 0 ≤ j, ∀ i. 1 < i ∧ i * i ≤ j → j % i ≠ 0 ⊦ j = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 18:
1 < i, 0 < i, i < j, 0 < j, 0 ≤ j, 2 ≤ j, ∀ i. 1 < i ∧ i * i ≤ j → j % i ≠ 0, ∃ i0. i * i0 = j ⊦ j = 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