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Proof step 2 from proof gcd-less (previous step)

i1 | i, i1 | j ⊦ 0 = i, i1 ≤ gcd(i, j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
gcd(i, j) < i1, i1 | j, i1 | i ⊦ 0 = i

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)