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Proof step 14 from proof div-minus-onediff-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 < j, i0 < ~1, i0 < 0 ⊦ 1 % j = 0, (i0 * j + 1) / j = i0 + 1, j = 0

Rule argument

Additional information

Used simplifier rules: numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)