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Proof step 26 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)) → (∃ m. n0 ≤ ii(m)) ∨ (∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)) ⊦ ∃ 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 27:
ii(m) ≤ m0, ∀ 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)) ⊦ ∀ n. ii(n) ≤ ii(n + 1), ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n), ∃ m1. m0 < ii(m1), ∃ m. ii(m) = m0
Proof step 29:
(∃ m. m0 ≤ ii(m)) ∨ (∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)), ii(m) ≤ m0, ∀ 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)) ⊦ ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n), ∃ m1. m0 < ii(m1), ∃ m. ii(m) = m0

Rule argument

subst: [ii, n0] → [ii, m0]

Additional information

Used simplifier rules: s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: ii, m0