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Proof step 12 from proof two-minus (previous step)

⊦ j * j + i * i = (i * (i - j) - j * (i - j)) + 2 * i * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ 2 * i * j + i * (i - j) = (j * j + i * i) + j * (i - j)

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; 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%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs