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

# σ = # σ0, (σ to n) + σ1 = σσ[n0], (σ0 to n) + σ2 = σσ[n1], ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, mon(σσ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0) ⊦ n0 = 0, σ to n = σ0 to n, σσ = [], # σ0 < ⌜n⌝

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:
# σ = # σ0, (σ0 to n) + σ2 = σσ[n1], (σ to n) + σ1 = σσ[n0], n1 < n0, ⌜n1⌝ < # σσ, ⌜n0⌝ < # σσ, mon(σσ), σσ[n1] ⊑ σσ[n0], ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ) ⊦ n0 = 0, σ to n = σ0 to n, σσ = [], # σ0 < ⌜n⌝

Rule argument

spec: <toplevel> name: mon-lem seq: mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ[m] ⊑ σσ[n] subst: [σσ, m, n] → [σσ, n1, n0]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s) Used terms: σσ, n1, n0