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Proof step 12 from proof four-square-odd-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≤ (2 * j)^2, abs(i6) ≤ j, abs(i5) ≤ j, abs(i4) ≤ j, abs(i3) ≤ j, 0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j, (2 * j + 1)^2 ≤ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j ^2, 0 < (2 * j)^2, (2 * j + 1)^2 ≤ (2 * j)^2 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2