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Proof step 10 from proof lagrange-lemma-smallest (previous step)

i1 ^2 + i2 ^2 + i3 ^2 + i4 ^2 = j1 * j, 1 < j, prime(j), 0 < j, 0 < j1, j1 < j, ∀ j0, i1, i2, i3, i4, j. i→n(j0) < i→n(j1) → i1 ^2 + i2 ^2 + i3 ^2 + i4 ^2 = j0 * j ∧ j0 < j ∧ 0 < j0 ∧ 0 < j ∧ prime(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)) ⊦ 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 11:
i1 ^2 + i2 ^2 + i3 ^2 + i4 ^2 = j1 * j, 1 < j, prime(j), 0 < j, 0 < j1, j1 < j, ∀ j0, i1, i2, i3, i4, j. i→n(j0) < i→n(j1) → i1 ^2 + i2 ^2 + i3 ^2 + i4 ^2 = j0 * j ∧ j0 < j ∧ 0 < j0 ∧ 0 < j ∧ prime(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)) ⊦ j = 2, i1 ^2 + i2 ^2 + i3 ^2 + i4 ^2 = j1 * j ∧ 0 < j1 ∧ j1 < j ∧ (∀ i3, i4, i5, i6, j2. 0 < j2 ∧ j2 < j1 → i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≠ j2 * 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: i1, i2, i3, i4, j1 abort: F discard: F computed: T

Additional information

Used terms: i1, i2, i3, i4, j1