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Proof step 3 from proof add-in (previous step)

∀ j, i0. i1 + j = i1 + i0 → j = i0, i1 +1 + j = i1 +1 + i0 ⊦ j = i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(i1 + 1) + j = (i1 + 1) + i0, ∀ j, i0. i1 + j = i1 + i0 → j = i0 ⊦ j = i0

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1;%add-a: ⊦ i + j + i0 = (i + j) + i0;