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Proof step 10 from proof lim-exists (previous step)

# σσ[n] = ∞, # σσ = ∞, mon(σσ) ⊦ σσ = [], ∃ σ1. σσ[n0] + σ1 = σσ[n]

User applied rule apply lemma

And obtained the following new goals

Proof step 11:
mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ[m] ⊑ σσ[n]
Proof step 12:
# σσ = ∞, # σσ[n] = ∞, mon(σσ) ⊦ σσ = [], ∃ σ1. σσ[n0] + σ1 = σσ[n], n0 ≤ n ∧ ⌜n⌝ < # σσ

Rule argument

spec: <toplevel> name: mon-lem seq: mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ[m] ⊑ σσ[n] subst: [σσ, m, n] → [σσ, n0, n]

Additional information

Used simplifier rules: prefix-def: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0); Used terms: σσ, n0, n