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Proof step 17 from proof four-square-odd-ls (previous step)

(2 * j + 1)^2 ≤ 4 * j ^2, i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2 ≤ 4 * 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 ⊦ 4 * j ^2 = (2 * j) * 2 * j, j = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: neq: ⊦ j ^2 ≠ j * j ↔ false; from specification int-square%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult