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Proof step 12 from proof div-minus-onediff (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
j0 < j, j0 < 0, i0 * j + j0 < j, i0 * j + j0 < 0, i0 < j + j0, i0 < j, i0 ≤ 0, (i0 * j + j0) + 1 < j, 0 < j + j0, j0 + 1 < j, 1 < j, 0 < j ⊦ j0 = 0, i0 = ((i0 * j + j0) + 1 - j) / j + 1, i0 * j = j * (i0 - 1) + j

Rule argument

Additional information

Used simplifier rules: add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs