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Proof step 10 from proof mod-pred-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(i1 * j + j0) + 1 = i0 * j + j1, j1 < j, j0 < j, 1 < j, 0 < j, 0 < j1, 0 ≤ j0, 0 ≤ i1, ~1 ≤ i1 * j + j0 ⊦ j1 = j0 + 1, j1 = 0

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i ≠ j → (i ≤ j ↔ 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% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)