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Proof step 25 from proof mod-ls-same-02 (previous step)

0 < i % j, i < i % j, j < 0, 0 ≤ i ⊦ j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 26:
⊦ j < 0 ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j) ∧ i0 ≤ 0)

Rule argument

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

Additional information

Used simplifier rules: ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i < 0 → (i * j < 0 ↔ 0 < j); (s)% ⊦ i ≤ j → (j < i ↔ false); (as)%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 Used terms: i, j, i0, j0