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Proof step 15 from proof div-minus-onediff-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j + (i % j + i0 % j) / j
Proof step 17:
i0 < 0, i0 < ~1, 0 < j ⊦ 1 % j = 0, (i0 * j + 1) / j = i0 + 1, j = 0, 0 ≤ i0 * j ∧ 0 ≤ 1 ∧ 0 < j

Rule argument

spec: <toplevel> 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: <all paths>

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% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: j, 1, i0 * j