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Proof step 5 from proof gcd-mult (previous step)

⊦ j = 0, i = 0, i1 = 0, gcd(i * i1, j * i1) = gcd(i, j) * abs(i1)

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 = 0, i = 0, i1 = 0, gcd(abs(i * i1), abs(j * i1)) = gcd(i, j) * abs(i1)

Rule argument

spec: int-gcd inst: name: gcd-abs-rew-02 seq: ⊦ gcd(i, j) = gcd(abs(i), abs(j)) subst: [j, i] → [j * i1, i * i1] flags: rewrite, current

Additional information

Used terms: j * i1, i * i1