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Proof step 8 from proof mod-succ-01 (previous step)

(i0 * j + j0) + 1 = i1 * j + j1, j0 < j, 0 < j, j1 < j, 0 ≤ i0, 0 ≤ j0, 0 ≤ j1, 0 ≤ i1, 0 ≤ (i0 * j + j0) + 1 ⊦ j0 + 1 = j, j1 = j0 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
(i0 * j + j0) + 1 = i1 * j + j1, j1 < j, 0 < j, j0 < j, 0 ≤ j0, 0 ≤ i0, ~1 ≤ i0 * j + j0, 0 ≤ i1, 0 ≤ j1 ⊦ j0 + 1 = j, j1 = j0 + 1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) 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% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)