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Proof step 2 from proof eq-one-neg (previous step)

⊦ i * j = ~1 ↔ abs(j) = 1 ∧ j + i = 0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
i * j = ~1 ⊦ abs(j) = 1 ∧ j + i = 0
Proof step 8:
abs(j) = 1 ∧ j + i = 0 ⊦ i * j = ~1

Rule argument

pos: right 1

Additional information