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Proof step 2 from proof lim-exists (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 3:
# σσ = ∞, mon(σσ) ⊦ σσ = [], ∃ σ. (∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ)) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])))
Proof step 128:
mon(σσ) ⊦ # σσ = ∞, σσ = [], ∃ σ. (∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ)) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])))

Rule argument

fma: # σσ = ∞

Additional information

Used formulas: # σσ = ∞