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Proof step 74 from proof lim-unique (previous step)

σ + σ1 = σσ[n], ⌜m⌝ = # σ, ⌜m0⌝ = # σ0, ⌜n⌝ < # σσ, m0 < m, mon(σσ), ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])), ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ0 to m1) + σ1 = σσ[n])) ⊦ m = m0, m = 0, σσ = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 75:
σ + σ1 = σσ[n], ⌜m⌝ = # σ, ⌜m0⌝ = # σσ[n] + # σ2, ⌜n⌝ < # σσ, m0 < m, mon(σσ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])), ∀ m1. ¬ m0 < m1 → (∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ1. (σσ[n] + σ2 to m1) + σ1 = σσ[n0])) ⊦ m = m0, m = 0, σσ = []

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream% ⊦ #(σ + σ1) = # σ + # σ1; (s)