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Proof step 6 from proof lim-exists (previous step)

# σσ[n] = ∞, # 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 exists right

And obtained the following new goals

Proof step 7:
# σσ[n] = ∞, # σσ = ∞, mon(σσ) ⊦ σσ = [], (∀ n0. ∃ σ1. σσ[n0] + σ1 = σσ[n]) ∧ (∀ m. ¬ # σσ[n] < ⌜m⌝ → (∃ n0, σ1. (σσ[n] to m) + σ1 = σσ[n0]))

Rule argument

pos: right 2 terms: σσ[n] abort: F discard: T computed: T

Additional information

Used terms: σσ[n]