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Proof step 55 from proof lim-unique (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 56:
# σ0 = ⌜n⌝ , ∞ = # σ, mon(σσ), ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ0 to m) + σ1 = σσ[n])), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ1. (σ to n + 1) + σ1 = σσ[n0]) ⊦ σσ = []

Rule argument

pos: left 4 terms: n + 1 abort: F discard: T computed: F

Additional information

Used terms: n + 1