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Proof step 132 from proof lim-exists (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 133:
# σσ = ⌜n⌝ , n = n0 +1, mon(σσ) ⊦ σσ = [], ∃ σ. (∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ)) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])))

Rule argument

var: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ # σσ = ⌜0⌝ ↔ σσ = []; (s)