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Proof step 1 from proof gcd-abs (conclusion)

⊦ abs(gcd(i, j)) = gcd(i, j)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: gcd-zero-02: ⊦ ¬ gcd(i, j) < 0;% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ abs(i) = i ↔ 0 ≤ i; (s)