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Proof step 28 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, i5 ^2 ≤ j ^2 ∧ i6 ^2 ≤ j ^2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ i ^2 ≤ j ^2 ↔ abs(i) ≤ abs(j)
Proof step 30:
(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

Rule argument

spec: int-square inst: name: le seq: ⊦ i ^2 ≤ j ^2 ↔ abs(i) ≤ abs(j) subst: [j, i] → [j, i5] flags: rewrite

Additional information

Used terms: j, i5