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Proof step 64 from proof two-even (previous step)

2 * i = 2 * j2 ^2 + (2 * i1)^2 + 4 * i1 + 1 + 2 * (2 * i1 + 1) * j2 ⊦ (2 * i1 ^2) * 2 + (2 * i1) * 2 = (2 * i1)^2 + 4 * i1, i = ((2 * i1 + 1) / 2)^2 + ((2 * j2 + 2 * i1 + 1) / 2)^2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
2 * i = 2 * j2 ^2 + (2 * i1)^2 + 4 * i1 + 1 + 2 * (2 * i1 + 1) * j2 ⊦ 4 * i1 ^2 = (2 * i1)^2, i = ((2 * i1 + 1) / 2)^2 + ((2 * j2 + 2 * i1 + 1) / 2)^2

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% ⊦ i + j = i + i0 ↔ j = i0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult