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Proof step 2 from proof k-lsrec (previous step)

i +1 = j ∨ i +1 = j0 ∧ i +1 < j ⊦ i ≠ j ∧ ¬ j < i

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: succ: ⊦ i ≠ i +1;%ls: ⊦ ¬ i +1 < i;%ls: ⊦ ¬ i +1 < i;%trans: ⊦ i < j ∧ j < j0 → i < j0;