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Proof step 1 from proof exrat-gcd1 (conclusion)

ri ≠ 0r ⊦ ∃ i, j. ri = rational(i, j) ∧ ri.num = i ∧ ri.denom = j ∧ 0 < j ∧ gcd(i, j) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ri = rational(0, 1), gcd(ri.num, ri.denom) = 1

Rule argument

Additional information

Used simplifier rules: zero-def: ⊦ 0r = rational(0, 1);%make: ⊦ rational(ri.num, ri.denom) = ri;%denom-pos-01: ⊦ ¬ rj.denom ≤ 0;% ⊦ ¬ i < j ↔ j ≤ i;