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

abs(rational(i, i0).num) ≠ 0 ∧ 0 < rational(i, i0).denom ⊦ abs(rational(i, i0).num) / gcd(abs(rational(i, i0).num), rational(i, i0).denom) = abs(rational(i, i0).num)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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