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Proof step 14 from proof div-minus-onediff-01 (previous step)

0 < j, i0 < ~1 ⊦ 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-add-rew seq: ⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j + (i % j + i0 % j) / j subst: [j, i0, i] → [j, 1, i0 * j] flags: rewrite, current

Additional information

Used simplifier rules: div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%mod-div-zero: ⊦ j ≠ 0 → (i % j) / j = 0;%one-02: j ≠ 0 ⊦ 1 / j = 1 ↔ j = 1;%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%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%zero-gels: ⊦ i < 0 ∧ 0 ≤ j → ¬ j < i; from specification int-abs Used terms: j, 1, i0 * j