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Proof step 7 from proof four-square-odd-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
(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

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ax-04: ⊦ i + 0 = i; from specification int-basic2%one: ⊦ 1 * i = i; from specification int-mult%div-mult-num: j ≠ 0, j0 % j = 0 ⊦ (i * (* j0)) / (* j) = i * j0 / j; from specification int-div%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%mod-two-02: ⊦ (i * 2) % 2 = 0; from specification int-div