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Proof step 3 from proof eq (previous step)

∀ j. i0 = i0 + j → j = 0, i0 +1 = i0 +1 + j ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i0 = j + i0, ∀ j. i0 = i0 + j → j = 0 ⊦ j = 0

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;%succ-in: ⊦ i +1 = j +1 ↔ i = j; from specification int-basic1