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Proof step 19 from proof div-eq-same-01 (previous step)

0 = i0 * i + j0, j = i0 + 1, abs(j0) < j, 0 < j, i < 0, j0 < j, j0 ≤ 0 ⊦ j = 0, j = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
0 = i0 * i + j0, i < 0, j0 < i0, 0 < i0, j0 ≤ 0, j0 ≤ i0, abs(j0) ≤ i0 ⊦ 0 = i0, ~1 = i0

Rule argument

Additional information

Used simplifier rules: forward-lels: ⊦ 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% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ 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% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)