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Proof step 121 from proof lim-exists (previous step)

# σσ = ∞, ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ[n] ⊑ σσ[n + 1] ⊦ 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]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 122:
# σσ = ∞, ∀ n0. ∃ σ1. σσ[n0] + σ1 = σσ[n0 + 1] ⊦ σσ = [], ∃ σ. (∀ n. ∃ σ1. σσ[n] + σ1 = σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n, σ1. (σ to m) + σ1 = σσ[n])), ∃ n. # σσ[n] = ∞, ∃ x. σσ[n + 1] = σσ[n] + ⌜x⌝

Rule argument

Additional information

Used simplifier rules: prefix-def: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0);%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%addeqinf-cancel: ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; from specification nat-inf%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; from specification nat-inf