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Proof step 3 from proof Lagrange-SumOf4Squares-primes (previous step)

prime(j), j ≠ 2 ⊦ ∃ i, i0, i1, i2, j0. i ^2 + i0 ^2 + i1 ^2 + i2 ^2 = j0 * j ∧ 0 < j0 ∧ j0 < j ∧ (∀ i3, i4, i5, i6, j1. 0 < j1 ∧ j1 < j0 → i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≠ j1 * j)

Lemma lagrange-lemma-smallest