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Proof step 8 from proof four-square-odd-ls (previous step)

(2 * j + 1)^2 ≤ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j, 0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, abs(i3) ≤ j, abs(i4) ≤ j, abs(i5) ≤ j, abs(i6) ≤ j ⊦ j = 0

User applied rule cut formula

And obtained the following new goals

Proof step 9:
i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≤ j ^2 + j ^2 + j ^2 + j ^2, (2 * j + 1)^2 ≤ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j, 0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, abs(i3) ≤ j, abs(i4) ≤ j, abs(i5) ≤ j, abs(i6) ≤ j ⊦ j = 0
Proof step 18:
(2 * j + 1)^2 ≤ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j, 0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, abs(i3) ≤ j, abs(i4) ≤ j, abs(i5) ≤ j, abs(i6) ≤ j ⊦ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≤ j ^2 + j ^2 + j ^2 + j ^2, j = 0

Rule argument

fma: i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≤ j ^2 + j ^2 + j ^2 + j ^2

Additional information

Used formulas: i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≤ j ^2 + j ^2 + j ^2 + j ^2