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Proof step 20 from proof lim-unique (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ¬ # σ0 < ⌜n⌝ → σ0 to n = σ1 to n)
Proof step 22:
#(((σ0 to n) + σ2) + σ3) = #((σ to n) + σ1), # σ = # σ0, n1 < n0, ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n0. ¬ #(((σ0 to n) + σ2) + σ3) < ⌜n0⌝ → ((σ0 to n) + σ2) + σ3 to n0 = (σ to n) + σ1 to n0 ⊦ n0 = 0, σ to n = σ0 to n, σσ = [], # σ0 < ⌜n⌝

Rule argument

spec: stream name: extensionality-to seq: ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ¬ # σ0 < ⌜n⌝ → σ0 to n = σ1 to n) subst: [σ1, σ0] → [(σ to n) + σ1, ((σ0 to n) + σ2) + σ3] flags: <all paths>

Additional information

Used terms: (σ to n) + σ1, ((σ0 to n) + σ2) + σ3