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Proof step 38 from proof lst-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
i +1 < j +1 ∨ j +1 < i +1 ⊦ j = i, i < j, j < i

Rule argument

Additional information

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