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Proof step 53 from proof lim-unique (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 54:
∞ = # σ, # σ0 = ⌜n⌝ , 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 = []

Rule argument

Additional information

Used simplifier rules: less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf