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Proof step 3 from proof weakmon-lem (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 4:
∀ 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)
Proof step 16:
∀ n. ii(n) ≤ ii(n + 1) ⊦ ∃ m. n ≤ ii(m), ∃ n0. ∀ m. n0 ≤ m → ii(m) = ii(n0), ∃ m, n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)

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)