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Proof step 11 from proof lrec (previous step)

∀ m. m +1 = n0 ∨ m +1 < n0 ↔ m < n0 ⊦ m +1 < n0, m = n0, m +1 = n0, m +1 < n0 ↔ m < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m < n0, ∀ m. m +1 = n0 ∨ m +1 < n0 ↔ m < n0 ⊦ m = n0, m +1 = n0, n0 = 0, m +1 < n0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0;