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Proof step 5 from proof gcd-num-one (previous step)

¬(i ≠ 0 ∧ 0 < i0) ⊦ rational(i, i0).num = 0, gcd(rational(i, i0).num, rational(i, i0).denom) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: numerator-zero: ⊦ rational(0, j).num = 0;%denominator-zero: ⊦ rational(0, j).denom = 1;% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ gcd(i, 1) = 1;% ⊦ ¬ i < j ↔ j ≤ i;