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Proof step 22 from proof make (previous step)

i ≠ 0 ∧ 1 ≠ 0, i ≠ 0 ⊦ i = 0, rational(i, 1).num = i ∧ abs(1) / gcd(i, 1) = 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: numerator-pos: ⊦ i ≠ 0 ∧ 0 < j → rational(i, j).num = i / gcd(i, j);% ⊦ i / 1 = i;% ⊦ gcd(i, 1) = 1;