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Proof step 4 from proof tot (previous step)

∀ n0. (∀ m. n0 = m ∨ n0 < m ∨ m < n0) → n0 +1 = m0 ∨ n0 +1 < m0 ∨ m0 < n0 +1, ∀ m. n0 = m ∨ n0 < m ∨ m < n0 ⊦ n0 +1 = m0 +1, n0 +1 < m0 +1, m0 +1 < n0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ m. n0 = m ∨ n0 < m ∨ m < n0, ∀ n. (∀ m. n = m ∨ n < m ∨ m < n) → n +1 = m0 ∨ n +1 < m0 ∨ m0 = n ∨ m0 < n ⊦ n0 = m0, n0 < m0, m0 < n0

Rule argument

Additional information

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