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Proof step 10 from proof gcd-divides-both (previous step)

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

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);%ler: ⊦ i ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%abs-03: ⊦ 0 ≤ abs(i); from specification int-abs