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Proof step 5 from proof add-succ-01 (previous step)

(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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: j, i0

Additional information

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