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Proof step 45 from proof lim-unique (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
# σ = # σ0, # σ1 + # σ3 = # σ2, σ to n = (σ0 to n) + σ2 to n, mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0) ⊦ σσ = [], σ to n = σ0 to n, # σ0 < ⌜n⌝

Rule argument

Additional information

Used simplifier rules: ler: ⊦ m ≤ m; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; from specification stream%to-to-eq: ¬ # σ < ⌜n⌝ ⊦ σ to n to n = σ to n; from specification stream