I 137f
Quantifiers/everyday language/Quine/Kaplan/Geach/Cresswell:
not 1st order: E.g. some critics only admire each other
2nd order:
(Eφ)(Exφx u (x)(φx > x is a critics) u (x)(y)((φx u x admires y) > (x ≠ y u φy))).
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 ≠ y ... ").
Better: "admire each other" is a predicate that is applied to pairs.
139
Correct: "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.
I 140
.. "admires another linguist" must be a predicate which is applied to all logicians. - This shows that quantification of higher level is required.
>
Second order logic.
Problem: this leads to the fact that the possibilities to have different ranges are restricted.
I 142
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.
>
Game-theoretical semantics.
CresswellVsHintikka: only higher order entities. 2nd order quantification due to reference to quantities.
I 156
Branching quantifiers/Booles/Cresswell: "for every A there is a B".
(x)(Ey)
(x = z ⇔ y = w) u (Ax > By)
(z)(Ew)
2nd order translation: EφEψ(x)(z)((x = z ⇔ φ(x) = ψ(z)) u (Ax > Bφ(x)).
Function/unique image/assignment/logical form/Cresswell: "(x = z ⇔ φ(x) = ψ (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.
>
Orderability.
