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Proof step 21 from proof lim-unique (previous step)

# σ = # σ0, #(((σ0 to n) + σ2) + σ3) = #((σ to n) + σ1) ∧ (∀ n0. ¬ #(((σ0 to n) + σ2) + σ3) < ⌜n0⌝ → ((σ0 to n) + σ2) + σ3 to n0 = (σ to n) + σ1 to n0), mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0) ⊦ σσ = [], σ to n = σ0 to n, n0 = 0, # σ0 < ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
# σ = # σ0, # σ2 + # σ3 = # σ1, mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ n2. ¬ ⌜n⌝ + # σ1 < ⌜n2⌝ → ((σ0 to n) + σ2) + σ3 to n2 = (σ to n) + σ1 to n2 ⊦ σσ = [], σ to n = σ0 to n, n0 = 0, # 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%add-cancel-01: ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; from specification nat-inf%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ; from specification stream%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream%size-append: ⊦ #(σ + σ1) = # σ + # σ1; from specification stream