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Proof step 136 from proof lim-exists (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 137:
# σσ = ⌜n0⌝ + ⌜1⌝ , mon(σσ), n ≤ n0 ⊦ σσ = [], ∃ σ1. σσ[n] + σ1 = σσ[n0], ∃ σ. (∀ n. ¬ n0 < n → (∃ σ1. σσ[n] + σ1 = σ)) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ¬ n0 < n ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])))

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat