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Proof step 5 from proof gcd-div-rest (previous step)

i = i1 * gcd(i, j) + j0, abs(j0) < gcd(i, j) ⊦ j0 = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ gcd(i, j) | i
Proof step 7:
i = i1 * gcd(i, j) + j0, abs(j0) < gcd(i, j), gcd(i, j) | i ⊦ j0 = 0

Rule argument

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

Additional information

Used terms: i, j