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Proof step 2 from proof ls-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 4:
j = i1 * i0 + j1 ∧ 0 ≤ j1 ∧ j1 < i0, i ≤ j, i0 ≤ j0, i1 < i / j0, 0 ≤ j, 0 ≤ i, 0 < j0, 0 < i0 ⊦ j0 = 0, i0 = 0

Rule argument

spec: int-div inst: name: div-elim-pos seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [i, j, i0, j0] → [j, i0, i1, j1] flags: rewrite, current

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: j, i0, i1, j1