errata p. 86, line 2 after "must" insert "neither be nor."
86 INTRODUCTION
Just as a class must not be capable of being or not being a member of itself; so a relation must not be referent or relatum with respect to itself. This turns out to be equivalent to the assertion that ϕ!(x, y) cannot significantly be either of the arguments x or y in ϕ!(x, y). This principle, again, results from the limitation to the possible arguments to a function explained at the beginning of Chapter II.
We may sum up this whole discussion on incomplete symbols as follows.
The use of the symbol" (’Ix) (</>x) " as if in "f(’Ix) (x)" it directly represented an argument to the function f is rendered possible by the theorems
b :. E! (’Ix) (</>x). :): (x) .fx. :) .f(’Ix) (cJ>x),
b : (1X) (x) = (’Ix) (-’frx) . :) .f(’Ix) (</>x) = .f(’Ix) (’o/x),
b : E! (’Ix) (x) . :). (’Ix) (x) = (’Ix) (x),
b : (’Ix) (cfJx) = (’Ix) (-’frx) . = . (1X) (vx) = (lX) (x),
b : (1X) (x) = (’Ix) (’o/x). (1X) (vx) = (1X) (Xx) . :) . (’Ix) (x) = (’Ix) (Xx).
The use of the symbol "f£ ( <px)" (or of a single letter, such as a, to represent such a symbol) as if, in "f {f£ (cf>x)}," it directly represented an argument a to a function fa, is rendered possible by the theorems
b : (a) . fa . :) . f {f£ ( x)},
b : f£ (cpa;) = 5J ( vx) . :) . f {f£ ( </>x)} - f f tc ( ’o/x)} ,
b . f£ (x) = f£ (x),
b : f£(cf>x)=f£ (vx). = . tc(’o/x) =tc (</>x),
b : fi (x) = 1£ (yx) . f£ (’o/x) = f£ (xx) . :) .1£ (</>x) = 5J (Xx).
Throughout these propositions the types must be supposed to be properly adjusted, where ambiguity is possible.
The use of the symbol" fiy {(x, y)} (or of a single letter, such as R, to represent such a symbol) as if, in "j{f£y (x, y)}," it directly represented an argument R to a function fR, is rendered possible by the theorems
b : (R) .fR . :) .f {f£y </> (x, y)},
b : 5Jp (x, y) = fiy V (x, y) . :) .1 {f£y (x, y)} = 1 {f£y ’0/ (x, y)},
b . 5Jy (x, y) = f£y (x, y),
b : feP cf> (x, y) = f£y V (x, y) . = . f£y V (x, y) = 5Jy </> (x, y),
b : fip </> (x, y) = f£p ’0/ (x, y) . 5Jy V (x, y) = f£y X (x, y) .
b . 5Jy </> (x, y) = f£y X (x, y).
Throughout these propositions the types must be supposed to be properly adjusted where ambiguity is possible.
