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Proof step 5 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 6:
∀ i, j. i + j + i1 = (i + j) + i1 ⊦ i + j + i1 = (i + j) + i1

Rule argument

Additional information

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