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Proof step 32 from proof weakmon-lem (previous step)

ii(m) ≤ ii(m1), m0 ≤ ii(m1), 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)) ⊦ ii(m) = m0, ∃ m. ii(m) = m0, ∃ m1. m0 < ii(m1), ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
ii(m) < m0, ii(m) < ii(m1), m0 ≤ ii(m1), ∀ 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)) ⊦ ii(m) = m0, ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n), ∃ m1. m0 < ii(m1), ∃ m. ii(m) = m0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)