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Proof step 14 from proof lim-unique (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 15:
mon(σσ), m ≤ n, ⌜n⌝ < # σσ ⊦ σσ[m] ⊑ σσ[n]
Proof step 16:
σσ[n1] ⊑ σσ[n0], # σ = # σ0, (σ to n) + σ1 = σσ[n0], (σ0 to n) + σ2 = σσ[n1], mon(σσ), ⌜n0⌝ < # σσ, ⌜n1⌝ < # σσ, n1 < n0, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ0) ⊦ σσ = [], σ to n = σ0 to n, n0 = 0, # σ0 < ⌜n⌝

Rule argument

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

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat Used terms: σσ, n1, n0