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Proof step 9 from proof mod-pred-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(i1 * j + j0) + 1 = i0 * j + j1, j0 < j, 0 ≤ j0, 0 ≤ i1, j1 < j, ~1 ≤ i1 * j + j0, 0 < j1, 1 < j, 0 < j ⊦ j1 = 0, j1 = j0 + 1

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs