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Proof step 2 from proof associativity (previous step)

∀ i, j. i * j * j1 = (i * j) * j1 ⊦ i * j * j1 +1 = (i * j) * j1 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ i, j. i * j * j1 = (i * j) * j1 ⊦ i * j * j1 = (i * j) * j1

Rule argument

Additional information

Used simplifier rules: dist-rsplit: ⊦ i * (j + j0) = i * j + i * j0;%one: ⊦ 1 * i = i;%commutativity: ⊦ i * j = j * i;%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2