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Proof step 11 from proof zero-04 (previous step)

j / i0 + 1 < i0, i1 + j / i0 < (i1 * i0 + j0) + j, i1 + j / i0 < j0, i1 + j / i0 < i1 * i0 + j0, (i1 + j / i0) + 1 < i0, 0 < i0, i1 + j / i0 < 0, j0 < i0, i1 + j / i0 < i0, j / i0 < 0, j / i0 < i1, j / i0 < (i1 * i0 + j0) + j, j / i0 < j0, j / i0 < i0, 0 ≤ i1, 0 ≤ (i1 * i0 + j0) + j, 0 ≤ j0 ⊦ 0 < i0 ∧ 0 ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
j < i1 * i0 + j0, 0 < i1 * i0 + j0, j < i1, j < i0, j + 1 < i0, j / i0 < i0, j / i0 < j0, j / i0 < (i1 * i0 + j0) + j, j / i0 < i1, j / i0 < 0, i1 + j / i0 < i0, j0 < i0, i1 + j / i0 < 0, 0 < i0, (i1 + j / i0) + 1 < i0, i1 + j / i0 < j0, i1 + j / i0 < (i1 * i0 + j0) + j, j / i0 + 1 < i0, j < 0, j < j0, j / i0 + 1 < i1 * i0 + j0, j / i0 < i1 * i0 + j0, (i1 + j / i0) + 1 < i1 * i0 + j0, i1 + j / i0 < i1 * i0 + j0, j + 1 < i1 * i0 + j0, 0 ≤ j0, 0 ≤ (i1 * i0 + j0) + j, 0 ≤ i1 ⊦

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < i + j ↔ 0 < j; (s)% ⊦ i +1 = i + 1; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1% ⊦ i ≤ j ↔ ¬ j < i; (ss)