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Proof step 1 from proof abs-pos (conclusion)

⊦ 0 < j → abs(rational(i, j)) = rational(abs(i), j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < j ⊦ abs(rational(i, j)) = rational(abs(i), j)

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)