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Proof step 1 from proof k-lsrec (conclusion)

⊦ i < j ↔ i = j -1 ∨ (∃ j0. i = j0 -1 ∧ j0 < j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i +1 = j ∨ i +1 = j0 ∧ i +1 < j ⊦ i ≠ j ∧ ¬ j < i

Rule argument

Additional information

Used simplifier rules: pred: ⊦ i = j -1 ↔ i +1 = j;%ls-succ-03: ⊦ i < j → (i +1 < j ↔ i +1 ≠ j);%l: ⊦ ¬ i < j ↔ ¬ ¬(i = j ∨ j < i);