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Proof step 2 from proof gcdex-smallest-pos (previous step)

⊦ j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0 ∧ 0 ≤ i0 ∧ (∀ i1. 0 ≤ i1 ∧ i1 < i0 → ¬(∃ j1. gcd(i, j) = i * i1 + j * j1))

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
j ≠ 0 ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0 ∧ 0 ≤ i0
Proof step 4:
∃ i0, j0. gcd(i, j) = i * i0 + j * j0 ∧ 0 ≤ i0 ⊦ j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0 ∧ 0 ≤ i0 ∧ (∀ i1. 0 ≤ i1 ∧ i1 < i0 → ¬(∃ j1. gcd(i, j) = i * i1 + j * j1))

Rule argument

spec: int-gcd inst: name: gcdex-pos seq: j ≠ 0 ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0 ∧ 0 ≤ i0 subst: [j, i] → [j, i] flags: current, []

Additional information

Used terms: j, i