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Proof step 8 from proof lim-exists (previous step)

# σσ = ∞, # σσ[n] = ∞, mon(σσ) ⊦ σσ = [], (∃ σ1. σσ[n0] + σ1 = σσ[n]) ∧ (∃ n0, σ1. (σσ[n] to m) + σ1 = σσ[n0])

User applied rule case distinction

And obtained the following new goals

Proof step 9:
# σσ = ∞, # σσ[n] = ∞, mon(σσ) ⊦ ∃ σ1. σσ[n0] + σ1 = σσ[n], σσ = []
Proof step 18:
# σσ = ∞, # σσ[n] = ∞, mon(σσ) ⊦ ∃ n0, σ1. (σσ[n] to m) + σ1 = σσ[n0], σσ = []

Rule argument

pos: right 2

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