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Proof step 23 from proof weakmon-lem (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
∀ 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), ¬ m0 < ii(m) ∧ (∃ m. ii(m) = m0)

Rule argument

Additional information

Used simplifier rules: s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2