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Proof step 8 from proof equal-minus-one (previous step)

~1 ^ j = ~1 ⊦ abs(j) % 2 = 1, j = 0, 0 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
j % 2 = 0, 1 / ~1 ^ ~ j = ~1, j < 0 ⊦

Rule argument

Additional information

Used simplifier rules: Pot-neg: ⊦ i < 0 → j ^ i = 1 / j ^ abs(i);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i < j → ¬ j < i; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ j < 0 → (j % 2 = ~1 ↔ j % 2 ≠ 0); (ss)% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)% ⊦ ~ i = j ↔ i + j = 0; (s)%j ≠ 0 ⊦ ~ i % j = ~(i % j); (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)