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Proof step 8 from proof gcd-divides-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
j | gcd(abs(i), abs(i0)), abs(i0) ≤ abs(i) ⊦ j | i ∧ j | i0

Rule argument

Additional information

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