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Proof step 118 from proof lim-exists (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 119:
# σσ = ∞, mon(σσ) ⊦ σσ = [], ∃ x. σσ[n + 1] = σσ[n] + ⌜x⌝ , ∃ n. # σσ[n] = ∞, ∃ σ. (∀ n. ∃ σ1. σσ[n] + σ1 = σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n, σ1. (σ to m) + σ1 = σσ[n]))

Proof step 118 from proof lim-exists (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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