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Proof step 64 from proof lim-unique (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
⌜m⌝ = # σ, (σ0 to m + 1) + σ1 = σσ[n], ⌜m⌝ < # σ0, mon(σσ), ⌜n⌝ < # σσ, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])) ⊦ ⌜m⌝ = # σ0, σσ = []

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%size-succ: ⊦ # σ < ⌜n⌝ + ⌜1⌝ ↔ ¬ ⌜n⌝ < # σ; from specification stream%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf