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Proof step 140 from proof lim-exists (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 141:
# σσ = ⌜n0⌝ + ⌜1⌝ , mon(σσ) ⊦ σσ = [], # σσ[n0] < ⌜m⌝ , ¬ n0 < n0 ∧ (∃ σ1. (σσ[n0] to m) + σ1 = σσ[n0]), ∃ n. ¬ n0 < n ∧ (∃ σ1. (σσ[n0] to m) + σ1 = σσ[n]), ∃ σ. (∀ n. ¬ n0 < n → (∃ σ1. σσ[n] + σ1 = σ)) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ¬ n0 < n ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])))

Rule argument

pos: right 3 terms: n0 abort: F discard: F computed: T

Additional information

Used terms: n0