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Proof step 6 from proof Lagrange-SumOf4Squares-primes (previous step)

j0 = 1, i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = j0 * j, j0 < j, prime(j), 0 < j0, 0 < j, 1 < 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. j = i ^2 + i0 ^2 + i1 ^2 + i2 ^2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
prime(i2 ^2 + i1 ^2 + i0 ^2 + i ^2), 1 < i2 ^2 + i1 ^2 + i0 ^2 + i ^2, 0 < i2 ^2 + i1 ^2 + i0 ^2 + i ^2 ⊦ i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = 2, ∃ i6, i5, i4, i3. i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = i6 ^2 + i5 ^2 + i4 ^2 + i3 ^2

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%num-between: i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult