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Proof step 2 from proof mod-nle (previous step)

0 < i, 0 ≤ j ⊦ j ≤ j % i ↔ j < i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs%zero-le: ⊦ 0 < i → (i * j ≤ 0 ↔ j ≤ 0); from specification int-mult%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-left: ⊦ 0 * i = 0; from specification int-mult%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%ls-add: ⊦ 0 ≤ j1 ∧ j1 < j → (i0 * j + j1 < j ↔ i0 ≤ 0); from specification int-mult%ler: ⊦ i ≤ i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: j, i, i0, j0