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Proof step 66 from proof euler-four-square-identity (previous step)

⊦ 2 * (i3 * j1) * i1 * j3 + 2 * (i4 * j2) * i1 * j3 + 2 * (i3 * j3) * i4 * j4 + 2 * ((i1 * j2) * i2 * j1 + (i1 * j2) * i3 * j4) + 2 * (i2 * j1) * i3 * j4 + 2 * (i4 * j1) * (i1 * j4 + i2 * j3) + 2 * (i1 * j4) * i2 * j3 = (j2 * i3 * 2) * i1 * j4 + (j4 * i2 * 2) * (i1 * j3 + i3 * j1) + 2 * ((i1 * j2 + i2 * j1) + i3 * j4) * i4 * j3 + 2 * (((i1 * j1) * i2 * j2 + (i1 * j1) * i3 * j3) + (i1 * j1) * i4 * j4)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%zero-left: ⊦ 0 * i = 0; from specification int-mult%numeq: ⊦ (* i2) * i + j = (* i1) * i + j0 ↔ (i2 - i1) * i + j = j0; from specification int-mult%eq-succ-01: ⊦ (i + i0) * j + j0 = (i + i1) * j + j1 ↔ i0 * j + j0 = i1 * j + j1; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult