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Proof step 18 from proof Lagrange-SumOf4Squares-primes (previous step)

i6 ^2 + i5 ^2 + i4 ^2 + i3 ^2 = j0 * j, i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = j0 * j, j0 / 2 < j, 0 < j0, 1 < j0, 0 < j0 / 2, even(i6 + i5), even(i4 + i3), ∃ j. j0 = j + j, j0 < j, prime(j), 0 < j, j0 / 2 < j0, 1 < j ⊦ ((i4 - i3) / 2)^2 + ((i4 + i3) / 2)^2 + ((i6 - i5) / 2)^2 + ((i6 + i5) / 2)^2 = j0 / 2 * j, j0 = 1, j = 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
i6 ^2 + i5 ^2 + i4 ^2 + i3 ^2 = (2 * j1) * j, i2 ^2 + i1 ^2 + i0 ^2 + i ^2 = (2 * j1) * j, prime(j), 0 < j, 1 < j, j1 < j, even(i6 + i5), even(i4 + i3), 2 * j1 < j, 0 < j1 ⊦ j = 2, ((i4 - i3) / 2)^2 + ((i4 + i3) / 2)^2 + ((i6 - i5) / 2)^2 + ((i6 + i5) / 2)^2 = j1 * j

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%ls-same-01: ⊦ 0 < i → (i < i * j ↔ 1 < j); from specification int-mult%nummult-ls-02: ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0); from specification int-div%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%neq-two: ⊦ 2 * i ≠ 1; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%div-mult-num: j ≠ 0, j0 % j = 0 ⊦ (i * (* j0)) / (* j) = i * j0 / j; from specification int-div%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult