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Proof step 7 from proof gcd-divides-both (previous step)

j | gcd(abs(i), abs(i0)) ⊦ j | i ∧ j | i0, abs(i) < abs(i0) ∧ j | gcd(abs(i), abs(i0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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