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Proof step 2 from proof weakmon-lem (previous step)

∀ n. ii(n) ≤ ii(n + 1) ⊦ ∃ m. n ≤ ii(m), ∃ n0. ∀ m. n0 ≤ m → ii(m) = ii(n0)

User applied rule cut formula

And obtained the following new goals

Proof step 3:
∃ m, n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n), ∀ n. ii(n) ≤ ii(n + 1) ⊦ ∃ m. n ≤ ii(m), ∃ n0. ∀ m. n0 ≤ m → ii(m) = ii(n0)
Proof step 15:
∀ n. ii(n) ≤ ii(n + 1) ⊦ ∃ m, n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n), ∃ m. n ≤ ii(m), ∃ n0. ∀ m. n0 ≤ m → ii(m) = ii(n0)

Rule argument

fma: ∃ m, n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)

Additional information

Used formulas: ∃ m, n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)