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Proof step 3 from proof denum-same (previous step)

¬(abs(rj.num) ≠ 0 ∧ rj.denom ≠ 0) ⊦ rational(abs(rj.num), rj.denom).denom = rj.denom

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: denom-not-zero: ⊦ ri.denom ≠ 0;%num-zero-denum-one: ⊦ rj.num = 0 → rj.denom = 1;%denominator-zero: ⊦ rational(0, j).denom = 1;% ⊦ abs(i) = 0 ↔ i = 0;