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Proof step 3 from proof gcd-less (previous step)

gcd(i, j) < i1, i1 | j, i1 | i ⊦ 0 = i

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j)

Rule argument

spec: <toplevel> name: gcd-max seq: ⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j) subst: [i0, i, j] → [i1, i, j]

Additional information

Used simplifier rules: ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: i1, i, j