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Proof step 11 from proof lim-exists (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# σσ[n] = ∞, # σσ = ∞, mon(σσ), n < n0 ⊦ σσ = [], n0 = 0, ∃ σ1. σσ[n0] + σ1 = σσ[n]

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf