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Proof step 1 from proof add-succ-01 (conclusion)

⊦ (i + j)+1 = i + i0 ↔ j +1 = i0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

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;