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Proof step 15 from proof two-minus (previous step)

⊦ 2 * (i0 + j) * j + (i0 + j) * i0 = (j * j + (i0 + j) * (i0 + j)) + j * i0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: dist-lsplit seq: ⊦ (i + i0) * j = i * j + i0 * j subst: [j, i0, i] → [j, j, i0] flags: rewrite

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%numeq-02: ⊦ (* i2) * i + j = i + j0 ↔ (i2 - 1) * i + j = j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult Used terms: j, j, i0