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Proof step 1 from proof equal-minus-one (conclusion)

⊦ i ≠ 0 ∨ 0 ≤ j → (i ^ j = ~1 ↔ i = ~1 ∧ j ≠ 0 ∧ abs(j) % 2 = 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i ≠ 0 ∨ 0 ≤ j ⊦ i ^ j = ~1 ↔ i = ~1 ∧ j ≠ 0 ∧ abs(j) % 2 = 1

Rule argument

Additional information