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Proof step 8 from proof div-minus-onediff (previous step)

0 < j, i0 * j + j0 < 0, i0 < j, i0 < j + j0, j0 < 0, 0 < j + j0, i0 ≤ 0 ⊦ i0 = (j * (i0 - 1) + j0 + 1) / j + 1, j0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i % j = 0 ∧ i < 0 ∧ i0 < 0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j
Proof step 10:
(j * (i0 - 1)) % j = 0, 0 < j + j0, j0 < 0, i0 < j + j0, i0 < j, i0 * j + j0 < 0, j * (i0 - 1) < 0, j0 + 1 < 0, 0 < j, i0 ≤ 0 ⊦ i0 = ((j * (i0 - 1)) / j + (j0 + 1) / j) + 1, j0 = 0

Rule argument

spec: <toplevel> name: div-add1-negneg seq: ⊦ i % j = 0 ∧ i < 0 ∧ i0 < 0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j subst: [j, i0, i] → [j, j0 + 1, j * (i0 - 1)] flags: <all paths>

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;% ⊦ i * 0 = 0; (s)%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%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)% ⊦ i + 0 = i; (s)% ⊦ i + (j - i) = j; (s) Used terms: j, j0 + 1, j * (i0 - 1)