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Proof step 4 from proof lagrange-lemma-smallest (previous step)

∃ i, i0, j0. 1 + i ^2 + i0 ^2 = j0 * j ∧ 0 < j0 ∧ j0 < j, 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
1 + i0 ^2 + i ^2 = j0 * j, j0 < j, prime(j), 0 < j0, 0 < j, 1 < 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)

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%transitivity: ⊦ i < i0 ∧ i0 < i1 → i < i1; (forward) from specification oelem[intlista]%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1