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Proof step 17 from proof div-minus-onediff-01 (previous step)

0 < j, i0 < ~1, i0 < 0 ⊦ j = 0, (i0 * j + 1) / j = i0 + 1, 1 % j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: div-add1-negpos-neg-01 seq: ⊦ i % j = 0 ∧ i + i0 < 0 ∧ i0 % j ≠ 0 ∧ i < 0 ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j + 1 subst: [j, i0, i] → [j, 1, i0 * j] flags: rewrite, current

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%div-eq-zero: ⊦ 0 < i ∧ 0 < j → (i / j = 0 ↔ i < j);%one-01: ⊦ i = 1 → j % i = 0;%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%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%ls-same-03: ⊦ i < 0 → (i < i * j ↔ j < 1); from specification int-mult%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: j, 1, i0 * j