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Proof step 1 from proof lim-unique (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
mon(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 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ0 to m) + σ1 = σσ[n])) ⊦ σ = σ0, σσ = []

Rule argument

Additional information

Used simplifier rules: prefix-def: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0);