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Proof step 4 from proof gcd-same (previous step)

⊦ gcd(abs(i), abs(i)) = abs(i), i = 0

User applied rule apply lemma

And obtained the following new goals

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

Rule argument

spec: <toplevel> name: gcd-div1 seq: ⊦ gcd(i, j) | i subst: [i, j] → [abs(i), abs(i)]

Additional information

Used terms: abs(i), abs(i)