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Proof step 5 from proof div-gcd-is-neg (previous step)

i0 < 0, 0 < i0 / gcd(i, i0) ⊦ gcd(i, i0) = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ 0 ≤ gcd(i, j)

Rule argument

spec: <toplevel> name: gcd-pos seq: ⊦ 0 ≤ gcd(i, j) subst: [i, j] → [i, i0]

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < 0 ∧ 0 < j → ¬ 0 < i / j; (s) Used terms: i, i0