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Proof step 5 from proof lagrange-lemma-smallest (previous step)

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)

User applied rule exists right

And obtained the following new goals

Proof step 6:
1 + i0 ^2 + i ^2 = j0 * j, j0 < j, prime(j), 0 < j0, 0 < j, 1 < j ⊦ j = 2, i ^2 + i0 ^2 + 1 ^2 + 0 ^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), ∃ 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

pos: right 2 terms: i, i0, 1, 0, j0 abort: F discard: F computed: F

Additional information

Used terms: i, i0, 1, 0, j0