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Proof step 7 from proof equal-minus-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
1 / ~1 ^ ~ j = ~1, j % 2 = 0, 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%ls-01: ⊦ i < j → ¬ j < i; 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%mod2-neg-odd: ⊦ j < 0 → (j % 2 ≠ ~1 ↔ j % 2 = 0); from specification int-div%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j); from specification int-div%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1