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Proof step 4 from proof gcd-zero-04 (previous step)

⊦ gcd(abs(i), 0) = abs(i)

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ gcd(i, j) | i
Proof step 6:
gcd(abs(i), 0) | abs(i) ⊦ gcd(abs(i), 0) = abs(i)

Rule argument

spec: int-gcd inst: name: gcd-div1 seq: ⊦ gcd(i, j) | i subst: [i, j] → [abs(i), 0] flags: current, []

Additional information

Used terms: abs(i), 0