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Proof step 4 from proof div-add1-negpos-neg-02 (previous step)

i % j = 0, 0 < j, i < 0, i + i0 < 0, 0 ≤ i0 ⊦ (i + i0) / j = i / j + (i0 + (j - 1)) / j, 1 < j ∧ 0 ≤ i0 ∧ i0 % j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 1 < j ∧ 0 < i ∧ i % j ≠ 0 → (i + (j - 1)) / j = i / j + 1
Proof step 6:
i % j = 0, i + i0 < 0, i < 0, 0 < j, 0 ≤ i0 ⊦ (i + i0) / j = i / j + (i0 + (j - 1)) / j, 1 < j ∧ 0 ≤ i0 ∧ i0 % j = 0, 1 < j ∧ 0 < i0 ∧ i0 % j ≠ 0

Rule argument

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

Additional information

Used simplifier rules: div-add1-negpos-neg-01: ⊦ i % j = 0 ∧ i + i0 < 0 ∧ i0 % j ≠ 0 ∧ i < 0 ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j + 1;%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) Used terms: j, i0