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Proof step 1 from proof gcd-one (conclusion)

⊦ gcd(i, j) = 1 ↔ ¬(∃ i1. i1 > 1 ∧ i1 | i ∧ i1 | j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ gcd(i, j) = 1 ↔ ¬(∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j)

Rule argument

Additional information

Used simplifier rules: gr-def: ⊦ i > j ↔ j < i; from specification int-abs