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Proof step 11 from proof add-div-mult (previous step)

j0 < j, 0 ≤ i0, 0 < j, 0 ≤ j0, 0 ≤ i1, (j0 + i0 % j) / j = 1 ∨ (j0 + i0 % j) / j = 0 ⊦ (i1 + i0 / j) + (j0 + i0 % j) / j = (i1 * j + j0 + i0) / j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: div-add1-pospos seq: ⊦ i % j = 0 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j subst: [j, i0, i] → [j, j0 + i0, i1 * j] flags: rewrite, current

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%zero-left: ⊦ 0 * i = 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%add-zero-rev: ⊦ 0 + i = i; 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%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%zero-gs: ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0; from specification int-abs Used terms: j, j0 + i0, i1 * j