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Proof step 4 from proof four-square-odd-ls (previous step)

∃ j. j0 = j + j + 1, abs(i3) ≤ j0 / 2, abs(i4) ≤ j0 / 2, abs(i5) ≤ j0 / 2, abs(i6) ≤ j0 / 2, j0 ^2 ≤ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j0, 0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ⊦ j0 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, abs(i3) ≤ (2 * j + 1) / 2, abs(i4) ≤ (2 * j + 1) / 2, abs(i5) ≤ (2 * j + 1) / 2, abs(i6) ≤ (2 * j + 1) / 2, (2 * j + 1)^2 ≤ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%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%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%two-zero: ⊦ 2 * i = 0 ↔ i = 0; from specification int-mult%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; 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%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult