40 1 Zum grammatischen Rahmen
Gelte die Behauptung fur Δ0, Δ1 ∈ FORM und sei Δ = r-Δ,√ ∈ JFORM. Dann ist [θ*1,
Θ1, [θ*0, θ0, Δ]] = [θ*1, θ1, [θ*0, θ0, r- Δ ]] = r-[θ*1, θ1, [θ*0, θ0, Δo]]π . Mit I.V. gilt [θ*b
Θ1, [θ*o, θo, Δo]] = [θ*o, θo, [θ*1, θ1, Δo]]. Also [θ*1, θb [θ*o, θ0, Δ]] = r-[θ*o, θ0, [θ*b θb
Δ Il = [θ*o, θo, [θ*1, θ1, ' Δ ]] = [θ*o, θo, [θ*1, θ1, Δ]]. Sei Δ = γ(Δo ψ Δ) ∈ JFORM.
Der Fall verlauft analog zum Negatorfall.
Sei Δ = rΠξΔ0^l ∈ QFORM. Angenommen ξ = θ0. Dann ist ξ ≠ θɪ und [θ*ι, θ1, [θ*0, θ0,
Δ]] = [θ*1, θ1, [θ*o, θo, r∏ξΔoπ]] = [θ*1, θ1, r∏ξΔoπ] = rΠξ[θ*1, θ1, ΔoΓ = [θ*o, θo,
rΠξ[θ*b θ1, Δo]T] = [θ*o, θo, [θ*1, θ1, FξΔ Ц = [θ*o, θo, [θ*1, θɪ, Δ]]. Angenommen ξ =
θ1. Dann ist ξ ≠ θo und [θ*b θɪ, [θ*o, θo, Δ]] = [θ*1, θɪ, [θ*o, θo, r∏ξΔoπ]] = [θ*1, θɪ,
rΠξ[θ*0, θo, Δo]π] = rΠξ[θ*0, θo, Δ ] = [θ*0, θo, r∏ξΔoπ] = [θ*0, θo, [θ*1, θ1, r∏ξΔoπ]] =
[θ*o, θo, [θ*1, θ1, Δ]]. Sei θo ≠ ξ ≠ θ1. Der Fall verlauft analog zum Negatorfall. ■
Theorem 1-26. Substitution in Substitutionsergebnissen
Wenn ζ ∈ VAR, θ', θ* ∈ GTERM und θ+ ∈ KONST ∪ PAR, dann:
(i) Wenn θ ∈ TERM, dann [θ', θ+, [θ*, ζ, θ]] = [[θ', θ+, θ*], ζ, [θ', θ+, θ]], und
(ii) Wenn Δ ∈ FORM, dann [θ', θ+, [θ*, ζ, Δ]] = [[θ', θ+, θ*], ζ, [θ', θ+, Δ]].
Beweis: Seien ζ ∈ VAR, θ', θ* ∈ GTERM und θ+ ∈ KONST ∪ PAR. Zu (i): Sei θ ∈
TERM. Der Beweis wird mittels Induktion uber den Termaufbau von θ gefuhrt. Sei θ ∈
ATERM. Sei weiter θ ∈ KONST ∪ PAR. Angenommen θ = θ+. Dann ist [θ', θ+, [θ*, ζ,
θ]] = [θ', θ+, θ] = θ'. Nun ist ζ ∉ TT(θ') ∈ GTERM und daher [θ', θ+, [θ*, ζ, θ]] = θ' = [[θ',
θ+, θ*], ζ, θ'] = [[θ', θ+, θ*], ζ, [θ', θ+, θ]]. Angenommen θ ≠ θ+. Dann ist [θ', θ+, [θ*, ζ, θ]]
= [θ', θ+, θ] = θ = [[θ', θ+, θ*], ζ, θ] = [[θ', θ+, θ*], ζ, [θ', θ+, θ]]. Sei schlieβlich θ ∈ VAR.
Angenommen θ = ζ. Dann ist [θ', θ+, [θ*, ζ, θ]] = [θ', θ+, θ*] = [[θ', θ+, θ*], ζ, θ] = [[θ', θ+,
θ*], ζ, [θ', θ+, θ]]. Angenommen θ ≠ ζ. Dann ist [θ', θ+, [θ*, ζ, θ]] = [θ', θ+, θ] = θ = [[θ',
θ+, θ*], ζ, θ] = [[θ', θ+, θ*], ζ, [θ', θ+, θ]].
Gelte die Behauptung nun fur {θ0, ., θr-1} ⊆ TERM und sei θ = rφ(θ0, ., θr-ι)^l ∈
FTERM. Dann ist [θ', θ+, [θ*, ζ, θ]] = [θ', θ+, [θ*, ζ, rφ(θo, ., θr-1)π]] = rφ([θ', θ+, [θ*, ζ,
θo]], ., [θ', θ+, [θ*, ζ, θr-1]])T Mit I.V. gilt [θ', θ+, [θ*, ζ, θi]] = [[θ', θ+, θ*], ζ, [θ', θ+, θi]]
fur alle i < r. Also [θ', θ+, [θ*, ζ, θ]] = rφ([[θ', θ+, θ*], ζ, [θ', θ+, θɑ]], ., [[θ', θ+, θ*], ζ, [θ',
θ+, θr-1]]Γ = [[θ', θ+, θ*], ζ, [θ', θ+, rφ(θo, ., θr-ʃ]] = [[θ', θ+, θ*], ζ, [θ', θ+, θ]].
Zu (ii): Sei Δ ∈ FORM. Der Beweis wird mittels Induktion uber den Formelaufbau von
Δ gefuhrt. Sei Δ = rΦ(θ0, . θr-ι)^l ∈ AFORM. Der Fall verlauft analog zum FTERM-Fall
unter Verwendung von (i).
More intriguing information
1. Convergence in TFP among Italian Regions - Panel Unit Roots with Heterogeneity and Cross Sectional Dependence2. BODY LANGUAGE IS OF PARTICULAR IMPORTANCE IN LARGE GROUPS
3. The Demand for Specialty-Crop Insurance: Adverse Selection and Moral Hazard
4. The name is absent
5. The open method of co-ordination: Some remarks regarding old-age security within an enlarged European Union
6. Trade Liberalization, Firm Performance and Labour Market Outcomes in the Developing World: What Can We Learn from Micro-LevelData?
7. The name is absent
8. The name is absent
9. Imputing Dairy Producers' Quota Discount Rate Using the Individual Export Milk Program in Quebec
10. Modelling the health related benefits of environmental policies - a CGE analysis for the eu countries with gem-e3