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Proof step 3 from proof add-succ-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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