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Proof step 3 from proof div-ls (previous step)

i0 % j = 0, i < i0, 0 < i0, 0 < j, j0 ≤ i / j, i0 / j ≤ j0, 0 ≤ i, i0 / j ≤ i / j ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 5:
0 < j, 0 < i0, i < i0, i0 / j ≤ i / j, 0 ≤ i, i0 / j ≤ j0, j0 ≤ i / j, ∃ j0. i0 = j0 * j ⊦ j = 0

Rule argument

spec: <toplevel> name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [j, i0] flags: <all paths>

Additional information

Used terms: j, i0