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Proof step 3 from proof equal-minus-one (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 4:
i ^ j = ~1, i ≠ 0 ∨ 0 ≤ j ⊦ i = ~1 ∧ j ≠ 0 ∧ abs(j) % 2 = 1
Proof step 50:
abs(j) % 2 = 1, i = ~1, i ≠ 0 ∨ 0 ≤ j ⊦ i ^ j = ~1, j = 0

Rule argument

pos: right 1

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