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Proof step 43 from proof lim-unique (previous step)

# σ = # σ0, # σ1 + # σ3 = # σ2, (σ to n) + σ1 to n = (σ0 to n) + σ2 to n, ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ) ⊦ σ to n = σ0 to n, σσ = [], # σ0 < ⌜n⌝ , #((σ to n) + σ1) < ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
# σ = # σ0, # σ1 + # σ3 = # σ2, (σ to n) + σ1 to n = (σ0 to n) + σ2 to n, ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, mon(σσ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0) ⊦ σ to n = σ0 to n, σσ = [], # σ0 < ⌜n⌝

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ ¬ n∞ + n∞1 < n∞1; (s)%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream%¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; (s)% ⊦ #(σ + σ1) = # σ + # σ1; (s)