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Proof step 2 from proof add-a (previous step)

∀ i, j. i + j + i1 = (i + j) + i1 ⊦ i + j + i1 +1 = (i + j) + i1 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ i, j. i + j + i1 = (i + j) + i1 ⊦ i + j + i1 = (i + j) + i1

Rule argument

Additional information

Used simplifier rules: add-succ: ⊦ i + j +1 = (i + j)+1;%succ-in: ⊦ i +1 = j +1 ↔ i = j; from specification int-basic1