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Proof step 27 from proof lst-02 (previous step)

j = i +1, i0 < i ∨ i < i0 ⊦ i0 +1 = j, i0 +1 < j, j < i0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
i0 < i ∨ i < i0 ⊦ i0 = i, i0 +1 < i +1, i +1 < i0 +1

Rule argument

Additional information

Used simplifier rules: succ-in: ⊦ i +1 = j +1 ↔ i = j;