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Proof step 4 from proof add-succ-02 (previous step)

∀ i. i +1 + j0 = (i + j0)+1 ⊦ i +1 + j0 -1 = (i + j0 -1)+1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ i. i +1 + j0 = (i + j0)+1 ⊦ (i + j0)+1 = i +1 + j0

Rule argument

Additional information

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