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Proof step 5 from proof gcd-divides-both (previous step)

j | gcd(i, i0) ⊦ j | i ∧ j | i0, 0 ≤ abs(i) ∧ abs(i) < abs(i0) ∧ abs(j) | gcd(abs(i), abs(i0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j | gcd(i, i0) ⊦ j | i ∧ j | i0, abs(i) < abs(i0) ∧ j | gcd(abs(i), abs(i0))

Rule argument

Additional information

Used simplifier rules: divides-abs-01: ⊦ abs(i) | j ↔ i | j;% ⊦ 0 ≤ abs(i); (s)