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Proof step 38 from proof lim-unique (previous step)

# σ = # σ0, # σ1 + # σ3 = # σ2, mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ n2. ¬ ⌜n⌝ + # σ2 < ⌜n2⌝ → ((σ to n) + σ1) + σ3 to n2 = (σ0 to n) + σ2 to n2 ⊦ σσ = [], σ to n = σ0 to n, # σ0 < ⌜n⌝

User applied rule all left

And obtained the following new goals

Proof step 39:
# σ = # σ0, # σ1 + # σ3 = # σ2, mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ¬ ⌜n⌝ + # σ2 < ⌜n⌝ → ((σ to n) + σ1) + σ3 to n = (σ0 to n) + σ2 to n ⊦ σσ = [], σ to n = σ0 to n, # σ0 < ⌜n⌝

Rule argument

pos: left 8 terms: n abort: F discard: T computed: F

Additional information

Used terms: n