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Proof step 5 from proof notneg-02 (previous step)

i ^ j + 1 < i, 0 < i, i ^ j < 0, i ^ j < i, 0 ≤ i ∧ 0 ≤ j → ¬ i ^ j < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i ^ ~ j + 1 = 0, i ^ ~ j < i, j < i, j + 1 < i, i ^ ~ j < 0, 1 / i ^ ~ j + 1 < i, 1 / i ^ ~ j < i, j < 0, 0 < i, j ≤ i ^ ~ j, 0 ≤ i ⊦

Rule argument

Additional information

Used simplifier rules: Pot-neg: ⊦ i < 0 → j ^ i = 1 / j ^ abs(i);%zero-02: ⊦ j < 0 → (i ^ ~ j = 0 ↔ i = 0);%zero-03: ⊦ j < 0 → 0 ^ ~ j = 0;% ⊦ i ≤ i; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < j + 1 ↔ i ≤ j; (s)%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%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%lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%mone-03: ⊦ i + 1 = 0 → i < 0; (forward) from specification int-basic2% ⊦ i +1 = i + 1; (s)% ⊦ i = ~1 ↔ i +1 = 0; (s)%j ≠ 0 ⊦ 1 / j < 0 ↔ j = ~1; (s)% ⊦ 0 - i = ~ i; (s)% ⊦ j < 0 → abs(j) = 0 - j; (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)