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Proof step 13 from proof four-square-odd-ls (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-square inst: name: le seq: ⊦ i ^2 ≤ j ^2 ↔ abs(i) ≤ abs(j) subst: [j, i] → [2 * j, 2 * j + 1] flags: rewrite

Additional information

Used simplifier rules: numabs-01: 0 ≤ j, 0 ≤ j0 ⊦ 0 < i → abs(i * (* j) + (* j0)) = i * j + j0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%numle-02: ⊦ (* i2) * i + j ≤ (* i1) * i ↔ (i2 - i1) * i + j ≤ 0; from specification int-mult%Mult-zero: ⊦ i * 0 = 0; from specification int-mult Used terms: 2 * j, 2 * j + 1