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Proof step 1 from proof gcdex-smallest-pos (conclusion)

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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information