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Proof step 30 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, 4 * j ^2 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 3 * j ^2 < i4 ^2 + i5 ^2 + i6 ^2 ⊦ j = 0, i4 ^2 ≤ j ^2 ∧ i5 ^2 + i6 ^2 ≤ 2 * j ^2, abs(i5) ≤ abs(j) ∧ i6 ^2 ≤ j ^2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
(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, 4 * j ^2 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 3 * j ^2 < i4 ^2 + i5 ^2 + i6 ^2, j ^2 < i6 ^2 ⊦ j = 0, i4 ^2 ≤ j ^2 ∧ i5 ^2 + i6 ^2 ≤ 2 * j ^2

Rule argument

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs