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Proof step 16 from proof negate (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
0 < j ⊦ rational(i, j).denom = rational(~ rational(i, j).num, rational(i, j).denom).denom

Rule argument

Additional information

Used simplifier rules: make: ⊦ rational(ri.num, ri.denom) = ri;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~ i + i = 0;