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Proof step 6 from proof mod-nls (previous step)

i = i1 + 1, 0 < i1 * j + i0, j < 0, i0 < i, 0 < i, 0 ≤ i0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 < i1 * j + i0, j < 0, i0 ≤ i1, 0 ≤ i1, 0 ≤ i0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ 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% ⊦ i < j + 1 ↔ i ≤ j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)