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Proof step 28 from proof weakmon-lem (previous step)

ii(m) ≤ m0, ii(n + 1) < ii(n), ∀ n. ii(n) ≤ ii(n + 1), ∀ n0, ii. n0 < m0 + 1 → (∀ n. ii(n) ≤ ii(n + 1)) → (∃ m. n0 ≤ ii(m)) ∨ (∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)) ⊦ ii(n) = 0, ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n), ∃ m1. m0 < ii(m1), ∃ m. ii(m) = m0

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n abort: F discard: T computed: F

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: n