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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ gcd(i, j) = gcd(abs(i), abs(j))
Proof step 7:
j | gcd(abs(i), abs(i0)) ⊦ j | i ∧ j | i0, abs(i) < abs(i0) ∧ j | gcd(abs(i), abs(i0))

Rule argument

spec: int-gcd inst: name: gcd-abs-rew-02 seq: ⊦ gcd(i, j) = gcd(abs(i), abs(j)) subst: [j, i] → [i0, i] flags: rewrite, current

Additional information

Used terms: i0, i