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Proof step 65 from proof lim-unique (previous step)

⌜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 = []

User applied rule all left

And obtained the following new goals

Proof step 66:
⌜m⌝ = # σ, (σ0 to m + 1) + σ1 = σσ[n], ⌜m⌝ < # σ0, mon(σσ), ⌜n⌝ < # σσ, ⌜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

pos: left 6 terms: n abort: F discard: T computed: T

Additional information

Used terms: n