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Proof step 4 from proof add-succ-01 (previous step)

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

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 5:
(j + i1)+1 = i0 + i1, (i1 + j)+1 = i1 + i1 → j +1 = i1, ∀ j, i0. (i1 + j)+1 = i1 + i0 → j +1 = i0 ⊦ j +1 = i0

Rule argument

pos: left 2 terms: j, i1

Additional information

Used terms: j, i1