September 2010
Intermediate to advanced
310 pages
12h 1m
German
Content preview from Handlungstheoretische Semantik
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Beweise D85
Für (*.1): Gegeben KR(X,P(s),f) und b P(s) und b ≠ X. Dann nach
D31(a) sowohl (α) P(X,T(X,f),T(b,f)) als auch () P(X,T(b,f),T(X,f)) als
auch (γ) P(X,T(X,f),T(b,f)), wegen P4.1 und P4.2 somit sowohl
G(X,(α)) als auch G(X,()) als auch G(X,(γ)), wegen TG1a also auch
G(X,(α) () (γ)), d.h. nach D31(a) also, da b beliebig,
G(X,KR(X,P(s),f)).
Für (*): Sei gegeben G(X,KR(X,P(s),f)), dann parallel wie soeben bei b
P(s) und b ≠ X zuerst G(X,(α) () (γ)) und daraus wieder mit TG1a
sowohl G(X,(α)) als auch G(X,()) Aus GGals auch G(X,(γ)), mit Hilfe
von P4.1 und P4.2 dann wieder (α) () (γ), womit, da b beliebig, nach
D31 schließlich KR(X,P(s),f).
Für (iii): Aus GG(P
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