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Proof step 58 from proof lim-unique (previous step)

# σ0 = ⌜n⌝ , (σ to n + 1) + σ1 = σσ[n0], ∞ = # σ, ⌜n0⌝ < # σσ, mon(σσ), ⌜n0⌝ < # σσ → (∃ σ1. σσ[n0] + σ1 = σ0), ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ0 to m) + σ1 = σσ[n])), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ) ⊦ σσ = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 59:
(σ to n + 1) + σ1 = σσ[n0], # σσ[n0] + # σ2 = ⌜n⌝ , ∞ = # σ, ⌜n0⌝ < # σσ, mon(σσ), ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σσ[n0] + σ2 to m) + σ1 = σσ[n])), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ) ⊦ σσ = []

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream% ⊦ #(σ + σ1) = # σ + # σ1; (s)