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Proof step 2 from proof gcd-negone-eq (
previous step
)
~1 = j2 / gcd(i2, j2) ⊦ gcd(i2, j2) = ~ j2
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
~1 = j2 / gcd(i2, j2) ⊦ j2 + gcd(i2, j2) = 0
Rule argument
Additional information
Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0; (s)