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Proof step 4 from proof div-add1-posneg-pos-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

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

Rule argument

spec: <toplevel> name: div-add-succ-neg seq: ⊦ 1 < j ∧ i < 0 ∧ 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-posneg-pos-01: ⊦ i % j = 0 ∧ 0 ≤ i + i0 ∧ i0 % j ≠ 0 ∧ 0 ≤ i ∧ i0 < 0 ∧ 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 - i0) = i + j - i0; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + (j - i) = j; (s)% ⊦ i + j = i + i0 ↔ j = i0; (s) Used terms: j, i0