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Proof step 23 from proof lim-unique (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
# σ = # σ0, # σ2 + # σ3 = # σ1, ((σ0 to n) + σ2) + σ3 to n = (σ to n) + σ1 to n, 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⌝

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-not-less: ⊦ ¬ n∞ + n∞1 < n∞1; from specification nat-inf