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Proof step 9 from proof abs-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 < j ⊦ j = 0, abs(rational(0, j)).num = 0

Rule argument

Additional information

Used simplifier rules: numerator-zero: ⊦ rational(0, j).num = 0;%denominator-zero: ⊦ rational(0, j).denom = 1;