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

abs(i) < gcd(abs(i), abs(j)) ⊦ i = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ gcd(i, j) | i
Proof step 7:
abs(i) < gcd(abs(i), abs(j)), gcd(abs(i), abs(j)) | abs(i) ⊦ i = 0

Rule argument

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

Additional information

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