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Proof step 3 from proof div-minus-onediff-01 (previous step)

i + 1 = i0 + j, i % j = 0, j + i ≤ 0, i < j, i + 1 < j, 0 < j, i < 0 ⊦ j = 1, i0 / j = i / j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 5:
j ≠ 0, i + 1 = i0 + j, ∃ j0. i = j0 * j, j + i ≤ 0, i < j, i + 1 < j, 0 < j, i < 0 ⊦ j = 1, i0 / j = i / j

Rule argument

spec: inst: name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used simplifier rules: add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs Used terms: j, i