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Proof step 4 from proof mod-nle-02 (previous step)

j0 < 0, i % j < 0, 0 < j, i < 0, i % j ≤ j0, j0 + j ≤ 0 ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)

Rule argument

spec: <toplevel> name: div-elim-negpos seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [i, j, i0, j1] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (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%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ i ≤ j → ¬ j < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) Used terms: i, j, i0, j1