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Proof step 6 from proof abs-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: num-same-01: ⊦ rational(abs(ri.num), ri.denom).num = abs(ri).num;%denum-same: ⊦ rational(abs(rj.num), rj.denom).denom = rj.denom;