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Proof step 7 from proof pot-zero-01 (previous step)

j < 0, 0 < i ⊦ 1 / i ^ abs(j) = 0 ↔ j < 0 ∧ 1 < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
j < i, j < 0, 0 < i, j + 1 < i ⊦ 1 / i ^ ~ j = 0 ↔ 1 < i

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%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%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs