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Proof step 11 from proof gcd-divides-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: divides-abs-01: ⊦ abs(i) | j ↔ i | j;%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%gcd-same-02: ⊦ gcd(abs(i), abs(i)) = abs(i);% ⊦ i ≤ i; (s)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 ≤ abs(i); (s)