1.2 Substitution
39
Sei Δ = rΠξΔoπ ∈ QFORM. Angenommen ξ = θ+. Dann ist [θ*, θ+, Δ] = [θ*, θ+,
'HξΔ I = rΠξΔoπ = [β, θ+, rΠξΔoπ] = [β, θ+, Δ]. Dann ist β ∉ TT([β, θ+, Δ]) = TT(Δ).
Also [θ*, θ+, Δ] = [β, θ+, Δ] = [θ*, β, [β, θ+, Δ]]. Angenommen ξ ≠ θ+. Der Fall verlauft
analog zum Negatorfall.
Zu (iii): Der Fall verlauft analog zum Negatorfall. ■
Theorem 1-25. Eine hinreichende Bedingung fur die Kommutativitat einer Substitution in
Termen und Formeln
Wenn θ*o, θ*ι ∈ GTERM, θo, θι ∈ ATERM, θo ≠ θɪ, θɪ ∉ TT(θ*o) und θo ∉ TT(θ*ι), dann:
(i) Wenn θ+ ∈ TERM, dann [θ*1, θ1, [θ*o, θo, θ+]] = [θ*o, θo, [θ*1, θ1, θ+]], und
(ii) Wenn Δ ∈ FORM, dann [θ*ι, θ1, [θ*o, θo, Δ]] = [θ*o, θo, [θ*ι, θι, Δ]].
Beweis: Seien θ*o, θ*1 ∈ GTERM, θo, θ1 ∈ ATERM, θo ≠ θ1, θ1 ∉ TT(θ*o) und θo ∉
TT(θ*1). Zu (i): Sei θ+ ∈ TERM. Der Beweis wird mittels Induktion uber den Termaufbau
von θ+ gefuhrt. Sei θ+ ∈ ATERM. Angenommen θ+ = θo. Dann ist θ+ ≠ θ1 und [θ*1, θ1,
[θ*o, θo, θ+]] = [θ*1, θ1, θ*o]. Weil θ1 ∉ TT(θ*o) gilt [θ*1, θ1, θ*o] = θ*o. Andererseits ist
[θ*o, θo, [θ*1, θ1, θ+]] = [θ*o, θo, θ+] = θ*o. Also [θ*1, θ1, [θ*o, θo, θ+]] = [θ*o, θo, [θ*1, θ1,
θ+]]. Sei nun θ+ ≠ θo. Angenommen θ+ = θ1. Dann ist [θ*1, θ1, [θ*o, θo, θ+]] = [θ*1, θ1, θ+]
= θ*1. Weil θo ∉ TT(θ*1) gilt [θ*o, θo, θ*1] = θ*1. Damit ist [θ*o, θo, [θ*1, θ1, θ+]] = [θ*o,
θo, θ*1] = θ*1. Also [θ*1, θ1, [θ*o, θo, θ+]] = [θ*o, θo, [θ*1, θ1, θ+]]. Angenommen θ+ ≠ θ1.
Dann ist [θ*1, θ1, [θ*o, θo, θ+]] = [θ*1, θ1, θ+] = θ+ und [θ*o, θo, [θ*1, θ1, θ+]] = [θ*o, θo, θ+]
= θ+. Also auch [θ*1, θ1, [θ*o, θo, θ+]] = [θ*o, θo, [θ*1, θ1, θ+]].
Gelte die Behauptung fur {θ'o, ..., θ'r-1} ⊆ TERM und sei θ+ = rφ(θ'o, ..., θ'r-1)^l ∈
FTERM. Dann ist [θ*ι, θ1, [θ*o, θo, θ+]] = [θ*ι, θι, [θ*o, θo, rφ(θ'o, ., θ'r-ι)π]] = rφ([θ*ι,
θι, [θ*o, θo, θ'o]], ., [θ*ι, θι, [θ*o, θo, θ'r-ι]])π. Mit I.V. gilt [θ*1, θι, [θ*o, θo, θ'i]] = [θ*o,
θo, [θ*ι, θι, θ'i]] fur alle i < r. Also [θ*ι, θι, [θ*o, θo, θ+]] = rφ([θ*o, θo, [θ*ι, θι, θ'o]], .,
[θ*o, θo, [θ*ι, θι, θ'r-ι]])π = [θ*o, θo, [θ*ι, θι, rφ(θ'o, . θ¼)π]] = [θ*o, θo, [θ*ι, θι, θ+]].
Zu (ii): Sei Δ ∈ FORM. Der Beweis wird mittels Induktion uber den Formelaufbau von
Δ gefuhrt. Sei Δ = rΦ(θ'o, . θ'r-ι)π ∈ AFORM. Dann ist [θ*ι, θι, [θ*o, θo, Δ]] = [θ*ι, θι,
[θ*o, θo, rΦ(θ'o, ., θ'r-1)π ]] = rΦ([θ*ι, θι, [θ*o, θo, θ'o]], ., [θ*1, θι, [θ*o, θo, θ'r-1]])π. Mit
(i) gilt [θ*1, θ1, [θ*o, θo, θ'i]] = [θ*o, θo, [θ*1, θ1, θ'i]] fur alle i < r. Also [θ*1, θ1, [θ*o, θo,
Δ]] = rΦ([θ*o, θo, [θ*ι, θι, θ'o]], ., [θ*o, θo, [θ*ι, θι, θ'r-1]])π = [θ*o, θo, [θ*ι, θι, rΦ(θ'o,
. θ'r-1)π ]] = [θ*o, θo, [θ*ι, θι, Δ]].
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