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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i0 * j + j0 < 0, i0 < j, i0 < j + j0, j0 < 0, 0 < j + j0, 0 < j, j0 < ~1, i0 ≤ 0 ⊦ (j0 + 1) / j = 0, j0 = 0

Rule argument

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i = i + j ↔ j = 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + (j - i) = j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * 0 = 0; (s)