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Proof step 3 from proof lrec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ m. m +1 = n0 ∨ m +1 < n0 ↔ m < n0 ⊦ m +1 = n0 ∨ m +1 < n0 ↔ m < n0, m = n0

Rule argument

Additional information

Used simplifier rules: inj-1: ⊦ n +1 = n0 +1 ↔ n = n0;%rax: ⊦ n0 < n +1 ↔ ¬ ¬(n0 = n ∨ n0 < n);