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Proof step 1 from proof gcd-less (conclusion)

⊦ i1 | i ∧ i1 | j ∧ 0 ≠ i → i1 ≤ gcd(i, j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i1 | i, i1 | j, gcd(i, j) < i1 ⊦ 0 = i

Rule argument

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs