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Proof step 12 from proof lim-exists (previous step)

# σσ = ∞, # σσ[n] = ∞, mon(σσ) ⊦ σσ = [], ∃ σ1. σσ[n0] + σ1 = σσ[n], n0 ≤ n ∧ ⌜n⌝ < # σσ

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# σσ = ∞, # σσ[n] = ∞, n < n0, mon(σσ) ⊦ n0 = 0, σσ = [], ∃ σ1. σσ[n0] + σ1 = σσ[n]

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ⌜m⌝ < ∞; (s)