[lojban] Re: A (rather long) discussion of {all}

[John E Clifford <lojban-out@lojban.org>]

And more

--- John E Clifford <clifford-j@sbcglobal.net> wrote:

> Trying again in a more straightforward way:
> 
> Singularist:
> 
> Domain D
> Masses M: all subsets of D with two or more members
> 
> Interpretation I
>  assigns to each name a member of  D u M
>  assigns to each predicate a set included in D u M
>  assigns to Y the relation between D u M and M that hold just in case the first relatum is
>  a member of or included in the second relatum. 
> 
> Assignment A assigns to each variable a member of D u M
> If A is an assignment, A(d/x) is an assignment just like A except for assigning d from 
> D u M to variable x in place of A(x)
> 
> R is a function from terms a to members of D u M, such that
> 	If a is a name, R(a) = I(a)
> 	If a is a variable, R(a) = A(a)
> 	If a = txF, R(a)  is some member d of D u M such that F is true for I and A(d/x)
> 
> A formula Pa is d-true for I and A iff  either I(a) e D and I(a) e I(P) or I(a) e M and for
>  every d e I(a) d e I(P).
> 
> A formula Pa is c-true for I and A iff I(a) e I(P).
> 
> A formula Pa is true for I and A iff it is either c-true or d-true
> 
> A formula ~F is true for I and A iff F is not true for I and A
> 
> A formula &FG is true for I and A iff both F and G are true for I and A
> 
> A formula ExF is true for I and A iff for some d e D u M, F is true for I and A(d/x)
> 
> 
> Pluralist:
> 
> Domain D
> 
> I is a relation whose first relatum is
> 	A name and whose second relatum is a member of d
> 	A predicate and whose second relatum is an n-place function over D into {0,1}
> 	Y and whose second relatum is an n+m-place function over D into {0,1}
> Such that each name is related to at least one member of D, each predicate is related to exactly
> one n-place function, for every n between 1 and the size of D, Y is related to exactly one
> function fnm for each n, m between 1 and the size of D, such that I(Y)(nm) (d1â?¦dn e1â?¦em)1
> iff 
> each ei is identical to one of the ds.
> 
> Since the array of functions for each predicate is unique as is the function for each number, we
> casn refer to the n-place function of a given predicate P as I(P)(n).
> 
> For convenience, we will abbreviate â??d1 â?¦ dn such that each di aIdiâ?? as I(a).  In the
> sequence d1 â?¦ dn it is understood that 1) no two items are identical and 2) the order of the
> items is not significant (the value of a function for d1â?¦dn in order is the same as the value
> for any permutation of that order).
> 
> We say â??d1â?¦dn numbers nâ??
> 
> A is a relation between variable and members of D
> A(d1â?¦dn/x) is a relation just like A except that x is related to each of d1â?¦ dn rather than
> to
> the things it is related to by A
> 
> We use A(x) analogously to I(a)
> 
> In the same vein we can define 
> 	R(a) = I(a) if a is a name
> 	R(a) = A(a) if a is a variable
> 	R(a) is some d1â?¦dn such that F is true for I and A(d1â?¦dn/x) if a = txF
Where R is a relation between terms and things in the domain such that 
	If a is a name then iff aRd, aId
	If a is a variable then iff aRd, aAd
	If a is txF, for some formula F and variable x, then there are d1,â?¦dn  such that F is true for
I and A(d2â?¦dn/x) and aRd iff d=di  (1 less  than or equal i less than or equal n)

> Pa is d-true for I and A iff for every d in R(a) I(P)(1)(d) = 1
> Pa is c-true for I and A iff R(a) numbers n and I(P)(n)(R(a)) = 1
> 
> F is true for I and A iff
> F = Pa and Pa is d-true for I and A or Pa is c-true for I and A.
> F = Yab and R(a) numbers n and R(b) numbers m and I(Y)(nm) (R(a)R(b)) =1
> F = ~G and G is not true for I and A
> F = &GH and both G and H are true for I and A
> F = ExG and, for some d1â?¦dn from D, G is true for I and A(d1â?¦dn/x)

[All this probably needs reference to R as well as I and A.] 
> 
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> 



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