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Proof step 1 from proof eq-one-neg (conclusion)

⊦ i * j = ~1 ↔ abs(i) = 1 ∧ i = ~ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ i * j = ~1 ↔ abs(j) = 1 ∧ j + i = 0

Rule argument

Additional information

Used simplifier rules: negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-neg: ⊦ abs(~ i) = abs(i); from specification int-abs