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Proof step 9 from proof div-abs-02 (previous step)

¬ ¬ j < 0, i * i0 = abs(j) ⊦ ∃ i0. i * i0 = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
j + i * i0 = 0, j < i * i0, j < 0, 0 < i * i0, j + 1 < i * i0 ⊦ ∃ i0. i * i0 = j

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ 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%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs