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Proof step 15 from proof div-add-succ (previous step)

(i0 * j) % j = 0, j0 < j, 0 < j, 0 ≤ j, 0 ≤ i0, 0 ≤ i0 * j + j0, 0 < i0 * j + j0, 1 < j, 0 < j0 ⊦ 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 seq: 0 < j, 0 ≤ i, i < j ⊦ i / j = 0 subst: [j, i] → [j, j0 - 1] flags: rewrite, current

Additional information

Used simplifier rules: sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ 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%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: j, j0 - 1