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Proof step 5 from proof less-mult-div-01 (previous step)

i ≤ (i * i1) / j, 0 < i, 0 < j, 0 < j1, 0 < (i * i1) / j, i1 ≤ 0, i1 < i, i1 < j, i1 < j1, i1 < (i * i1) / j ⊦

User applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)

Rule argument

spec: int-div inst: 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 * i1, j, i0, j0] flags: current, []

Additional information

Used simplifier rules: div-zero-05: ⊦ i < j ∧ 0 < j ∧ 0 ≤ i → i / j = 0;%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%zero-left: ⊦ 0 * i = 0; from specification int-mult Used terms: i * i1, j, i0, j0