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Proof step 13 from proof lim-exists (previous step)

# σσ = ∞, # σσ[n] = ∞, n < n0, mon(σσ) ⊦ n0 = 0, σσ = [], ∃ σ1. σσ[n0] + σ1 = σσ[n]

User applied rule exists right

And obtained the following new goals

Proof step 14:
# σσ[n] = ∞, # σσ = ∞, n < n0, mon(σσ) ⊦ σσ[n0] + [] = σσ[n], n0 = 0, σσ = []

Rule argument

pos: right 3 terms: [] abort: F discard: T computed: F

Additional information

Used terms: []