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Proof step 3 from proof ls-succ (previous step)

⊦ i +1 = j +1 ∨ i +1 < j +1 ↔ ¬ j < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ i = j ∨ i +1 < j +1 ↔ ¬ j < i

Rule argument

Additional information

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