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Proof step 9 from proof lim-unique (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# σ = # σ0, (σ to n) + σ1 = σσ[n0], ⌜n0⌝ < # σσ, mon(σσ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ m. ¬ # σ0 < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ0 to m) + σ1 = σσ[n])) ⊦ σ to n = σ0 to n, σσ = [], # σ0 < ⌜n⌝

Proof step 9 from proof lim-unique (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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