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Proof step 57 from proof lim-unique (previous step)

# σ0 = ⌜n⌝ , (σ to n + 1) + σ1 = σσ[n0], ∞ = # σ, ⌜n0⌝ < # σσ, mon(σσ), ∀ 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 58:
# σ0 = ⌜n⌝ , (σ to n + 1) + σ1 = σσ[n0], ∞ = # σ, ⌜n0⌝ < # σσ, mon(σσ), ⌜n0⌝ < # σσ → (∃ σ1. σσ[n0] + σ1 = σ0), ∀ m. ¬ n < m → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ0 to m) + σ1 = σσ[n])), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ) ⊦ σσ = []

Rule argument

pos: left 7 terms: n0 abort: F discard: T computed: T

Additional information

Used terms: n0