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Proof step 8 from proof abs (previous step)

j < 0, i < 0 ⊦ i / abs(j) + abs(i / j) = 0, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
i < 0, j < 0, i / j < 0 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ abs(i) = i ↔ 0 ≤ i; (s)% ⊦ i + 0 = i; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + ~ j = i - j; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)