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Proof step 135 from proof lim-exists (previous step)

# σσ = ⌜n0⌝ + ⌜1⌝ , mon(σσ) ⊦ σσ = [], ∃ σ. (∀ n. ¬ n0 < n → (∃ σ1. σσ[n] + σ1 = σ)) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ¬ n0 < n ∧ (∃ σ1. (σ to m) + σ1 = σσ[n]))), (¬ n0 < n → (∃ σ1. σσ[n] + σ1 = σσ[n0])) ∧ (¬ # σσ[n0] < ⌜m⌝ → (∃ n. ¬ n0 < n ∧ (∃ σ1. (σσ[n0] to m) + σ1 = σσ[n])))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 3

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