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Proof step 2 from proof lrec (previous step)

⊦ m +1 = n ∨ m +1 < n ↔ m < n

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ m. m +1 = n0 ∨ m +1 < n0 ↔ m < n0 ⊦ m +1 = n0 +1 ∨ m +1 < n0 +1 ↔ m < n0 +1

Rule argument

term: n

Additional information

Used simplifier rules: gen: ⊦ ;%disj: ⊦ 0 ≠ n +1;%less-1: ⊦ ¬ n < 0;