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Proof step 4 from proof lim-exists (previous step)

# σσ = ∞, mon(σσ) ⊦ σσ = [], ∃ σ. (∀ n. ∃ σ1. σσ[n] + σ1 = σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n, σ1. (σ to m) + σ1 = σσ[n]))

User applied rule cut formula

And obtained the following new goals

Proof step 5:
∃ n. # σσ[n] = ∞, # σσ = ∞, mon(σσ) ⊦ σσ = [], ∃ σ. (∀ n. ∃ σ1. σσ[n] + σ1 = σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n, σ1. (σ to m) + σ1 = σσ[n]))
Proof step 19:
# σσ = ∞, mon(σσ) ⊦ ∃ n. # σσ[n] = ∞, σσ = [], ∃ σ. (∀ n. ∃ σ1. σσ[n] + σ1 = σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n, σ1. (σ to m) + σ1 = σσ[n]))

Rule argument

fma: ∃ n. # σσ[n] = ∞

Additional information

Used formulas: ∃ n. # σσ[n] = ∞