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Proof step 3 from proof gcd-one (previous step)

gcd(i, j) = 1, ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
gcd(i, j) = 1, 0 < i0, 1 < i0, i0 | i, i0 | j ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1