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Proof step 72 from proof lim-unique (previous step)

⌜m0⌝ = # σ0, ⌜m⌝ = # σ, σ + σ1 = σσ[n], m0 < m, mon(σσ), ⌜n⌝ < # σσ, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ0 to m1) + σ1 = σσ[n])), ∀ 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 = 0, m = m0, σσ = []

User applied rule all left

And obtained the following new goals

Proof step 73:
⌜m0⌝ = # σ0, ⌜m⌝ = # σ, σ + σ1 = σσ[n], m0 < m, mon(σσ), ⌜n⌝ < # σσ, ∀ m1. ¬ m0 < m1 → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ0 to m1) + σ1 = σσ[n])), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])) ⊦ m = 0, m = m0, σσ = []

Rule argument

pos: left 9 terms: n abort: F discard: T computed: T

Additional information

Used terms: n