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Proof step 1 from proof gcd-negone-eq (conclusion)

⊦ ~1 = j2 / gcd(i2, j2) → gcd(i2, j2) = ~ j2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
~1 = j2 / gcd(i2, j2) ⊦ j2 + gcd(i2, j2) = 0

Rule argument

Additional information

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