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Proof step 1 from proof weakmon-lem (conclusion)

∀ n. ii(n) ≤ ii(n + 1) ⊦ (∀ n. ∃ m. ii(m) ≥ n) ∨ (∃ n. ∀ m. m ≥ n → ii(m) = ii(n))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ n. ii(n) ≤ ii(n + 1) ⊦ ∃ m. n ≤ ii(m), ∃ n0. ∀ m. n0 ≤ m → ii(m) = ii(n0)

Rule argument

Additional information

Used simplifier rules: ge: ⊦ n ≥ m ↔ m ≤ n; from specification nat