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Proof step 20 from proof four-square-odd-ls (previous step)

¬(i3 ^2 ≤ j ^2 ∧ i4 ^2 + i5 ^2 + i6 ^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 ⊦ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≤ j ^2 + j ^2 + j ^2 + j ^2, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
i ≤ j, i0 ≤ j0 ⊦ i + i0 ≤ j + j0
Proof step 22:
¬(i4 ^2 ≤ j ^2 ∧ i5 ^2 + i6 ^2 ≤ j ^2 + j ^2), ¬(i3 ^2 ≤ j ^2 ∧ i4 ^2 + i5 ^2 + i6 ^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 ⊦ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≤ j ^2 + j ^2 + j ^2 + j ^2, j = 0
Proof step 36:
i4 ^2 ≤ j ^2 ∧ i5 ^2 + i6 ^2 ≤ j ^2 + j ^2, ¬ i3 ^2 ≤ j ^2 ∨ ¬ true, (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

spec: int-abs inst: name: ple seq: i ≤ j, i0 ≤ j0 ⊦ i + i0 ≤ j + j0 subst: [j0, j, i0, i] → [j ^2 + j ^2, j ^2, i5 ^2 + i6 ^2, i4 ^2] flags: rewrite

Additional information

Used terms: j ^2 + j ^2, j ^2, i5 ^2 + i6 ^2, i4 ^2