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Proof step 27 from proof weakmon-lem (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 28:
ii(m) ≤ m0, ∀ n. ii(n) ≤ ii(n + 1), ∀ n0, ii. n0 < m0 + 1 → (∀ n. ii(n) ≤ ii(n + 1)) → (∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)) ∨ (∃ m. n0 ≤ ii(m)) ⊦ ∀ n. ii(n) ≤ ii(n + 1), ∃ m. ii(m) = m0, ∃ m1. m0 < ii(m1), ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)
Proof step 30:
ii(m) ≤ m0, ∀ n. ii(n) ≤ ii(n + 1), (∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)) ∨ (∃ m. m0 ≤ ii(m)), ∀ n0, ii. n0 < m0 + 1 → (∀ n. ii(n) ≤ ii(n + 1)) → (∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)) ∨ (∃ m. n0 ≤ ii(m)) ⊦ ∃ m. ii(m) = m0, ∃ m1. m0 < ii(m1), ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)

Rule argument

subst: [n0, ii] → [m0, ii]

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m0, ii