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Proof step 34 from proof weakmon-lem (previous step)

ii(m) < ii(m1), ii(m) < m0, 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, m0 < ii(m1), ∃ 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 35:
ii(m) < ii(m1), ∀ n. ii(n) ≤ ii(n + 1), ∀ n0, ii0. n0 < ii(m1) + 1 → (∀ n. ii0(n) ≤ ii0(n + 1)) → (∃ n. (∀ m. ii0(m) ≤ n) ∧ (∃ m. ii0(m) = n)) ∨ (∃ m. n0 ≤ ii0(m)) ⊦ ∃ m. ii(m) = ii(m1), ∃ m. ii(m1) < ii(m), ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)