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Proof step 3 from proof modlsp-03 (previous step)

j * i + j0 < j * i0 + j1, 0 ≤ j0, j0 < j, abs(j1) < j ⊦ j * i + (j0 - j1) < j * i0 ∧ abs(j0 - j1) < j, i ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs