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

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

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-gcd inst: name: gcd-divides-both-h seq: 0 ≤ i, i < j, j0 | gcd(i, j) ⊦ j0 | j ∧ j0 | i subst: [i, j, j0] → [abs(i0), abs(i), abs(j)] flags: current, []

Additional information

Used simplifier rules: divides-abs: ⊦ abs(i) | abs(j) ↔ i | j; Used terms: abs(i0), abs(i), abs(j)