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Proof step 2 from proof div-zero-01 (previous step)

i < j, i < 0, 0 < j + i, i + 1 < j, 1 < j, 0 < j ⊦ i / j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-elim2 seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j) subst: [i, j, i0, j0] → [i, j, i0, j0] flags: rewrite, current

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2%ls-add-both: ⊦ i < j → i1 + i < i1 + j; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%add-less: ⊦ 0 < i → i1 < i1 + i; from specification int-basic2%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%zero-gr-add: ⊦ 0 < j → (0 < i * j + j ↔ 0 ≤ i); from specification int-mult%lels-02: ⊦ i ≠ j → (i ≤ j ↔ 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%modzero: ⊦ 0 < i ∧ abs(j0) < j → ¬ i * j + j0 < 0; from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: i, j, i0, j0