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Proof step 3 from proof gcd-div-gcd-pos (previous step)

0 < i, 0 < j ⊦ ∀ i1. ¬(i1 > 1 ∧ i1 | i / gcd(i, j) ∧ i1 | j / gcd(i, j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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%gr-def: ⊦ i > j ↔ j < i; from specification int-abs