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Proof step 10 from proof div-minus-onediff (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-zero-02 seq: abs(i) < abs(j) ⊦ i / j = 0 subst: [j, i] → [j, j0 + 1] flags: rewrite, current

Additional information

Used simplifier rules: abs-05: ⊦ i < 0 → abs(i + 1) = abs(i) - 1; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; from specification int-basic2 Used terms: j, j0 + 1