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Quantifiers/everyday language/Quine/Kaplan/Geach/Cresswell: not 1st order: E.g. some critics only admire each other -2nd order: (Ej)(Exjx u (x)( jx > x is a critics) u (x)(y)(( jx u x admires y) > (x ungl y u jy))) - that is not equivalent to any 1st order sentence - involves plural noun phrases (plural quantification). - The following is not correct: "two Fs are G" - one would have to assume that "admire" should be valid in both directions - (then x is a K u y is a K u x ungl y ... ") - better: "admire each other" is a predicate that is applied to pairs.
Right: "Smart and Armstrong are present" for "S. is a and A is a". - Problem: "King and Queen are a lovable couple", then "The King is an adorable ..." analog: E.g. "similar", e.g. "lessen". Solution/Cresswell: applying predicate to quantities.
.. "admires another linguist" must be a predicate which is applied to all logicians. - This shows that quantification of higher level is required. - Problem: this leads to the fact that the possibilities to have different ranges are restricted.
Higher order quantifiers/plural quantifiers/Boolos: Thesis: these do not have to go via set theoretical entities, but can simply be interpreted as semantically primitive. ((s) basic concept) - Cresswell: perhaps he is right) - Hintikka: game theory - CresswellVsHintikka: only higher order entities. 2nd order quantification due to reference to quantities.
Branched quantifiers/Booles/Cresswell: "for every A there is a B".
(x = z ⇔ y = w) u (Ax > By)
2nd order translation: EfEg(x)(z)((x = z ⇔ f(x) = g(z)) u (Ax > Bf(x)) - Function/unique image/assignment/logical form/Cresswell: "(x = z ⇔ f (x) = g (z)" says that the function is 1: 1. - Generalization/Cresswell: If we replace W, C, A, B, and R by predicates that are true of all, and Lxyzw by Boolos ((x = z ⇔ y = w) u Ax> By) we have a proof of non-orderability of 1st order.
M. J. Cresswell
Semantical Essays (Possible worlds and their rivals) Dordrecht Boston 1988
M. J. Cresswell
Structured Meanings Cambridge Mass. 1984