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Proof step 15 from proof lim-exists (previous step)

# σσ = ∞, # σσ[n] = ∞, n < n0, mon(σσ) ⊦ σσ[n0] = σσ[n], n0 = 0, σσ = []

User applied rule apply lemma

And obtained the following new goals

Proof step 16:
mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ[m] ⊑ σσ[n]
Proof step 17:
# σσ[n] = ∞, # σσ = ∞, n < n0, mon(σσ), σσ[n] ⊑ σσ[n0] ⊦ σσ[n0] = σσ[n], n0 = 0, σσ = []

Rule argument

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

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)% ⊦ ⌜m⌝ < ∞; (s) Used terms: σσ, n, n0