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Proof step 6 from proof negate (previous step)

0 < j ⊦ ~ rational(i, j).num = (~ rational(i, j)).num ∧ rational(~ i, j).denom = (~ rational(i, j)).denom

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 < j ⊦ rational(i, j).num + (~ rational(i, j)).num = 0 ∧ rational(~ i, j).denom = (~ rational(i, j)).denom

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0;