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Proof step 3 from proof div-gcd-neg-not-zero (previous step)

i0 / gcd(i, i0) = 0, i0 < 0 ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ j ≠ 0 → (abs(i) < gcd(i, j) ↔ i = 0)
Proof step 5:
i0 / gcd(i, i0) = 0, i0 < 0, i ≠ 0 → (abs(i0) < gcd(i0, i) ↔ i0 = 0) ⊦

Rule argument

spec: <toplevel> name: abs-gcd-not-less seq: ⊦ j ≠ 0 → (abs(i) < gcd(i, j) ↔ i = 0) subst: [j, i] → [i, i0]

Additional information

Used terms: i, i0