Russell’s Paradox Solved!
by Dennis J. Darlaznd
June 12, 2007
· There are just two primitive types of symbols, nouns and verbs.
· The logical connectives (not, or, etc.) and quantifiers and variables are not symbols.
· Nouns may not be used as predicates.
· Any verb can have a name, but that name is a noun, not a verb.
· Nouns may also have a name, another noun.
· Verbs may only occur as predicates, or the second operand of a belief_r relation(or other propositional attitude) , or the, third operand of a symbol1_r relation. (Or (in prolog) the first operand of apply).
Russell defines two functions as formally equivalent when they when they are equivalent with every possible argument. (Principia Mathematica, p. 72.)
A function is extensional when its truth-value is the same as with any formally equivalent argument. (Principia Mathematica, p. 72)
A function is intensional whan it is not extensional. (Principia Mathematica, p. 73.)
An example of an intensional function is:
‘A believes that ‘x is a man’ of B”.
On my definition it would be belief_r(A,’man’,B,now)
Now Russell says ‘featherless biped’ is formally equivalent to ‘man’.
But we may not have belief_r(A,’featherless biped’,B,now).
So belief_r is an intensional function of ‘man’ (i.e. is a man.)
Russell defines h{ z : f(z) } as ($g) ( ( (x) (f(x) ó g!(x) ) & h(g ! (x)), (I’ve had to change the notation see page 76)
Russell defines
x e g!(x) as g!(x)
It follows from Russell’s definitions
x e {z : f(z)} means ($g) (y)( (f(y) ó g!(y)) & g!(x))
Lets consider
α = {z : z ~e z}
x e α
iff α!(x)
and it is true iff
($g)(((x)((z ~e z) (x) ó g!(x)) & (z ~e z)(x) ---- here (z ~e z) is the predicate true of things not members of themselves.
Another e appears so there is nothing that α can be used to abbreviate! Since class membership is an abbreviation it cannot be used to define a class! There is nothing to stop the recursion.
· The class notation in Principia Mathematica is just abbreviations which may be used to make complex propositions seem more familiar.
· All propositions of Principia Mathematica are finite. (Although they may say something about something that is infinite).
· When Russell’s rule for determining what the proposition using class symbols are an abbreviated proposition of are applied to ‘the class not a member of itself’, the expansion goes on without end.
· There is no finite proposition of Principia Mathematica , not using class notation, that ‘the class not a member of itself’ can be used to abbreviate.
· The class notation does not add anything – the abbreviations are just a convenience.
· The symbols ‘the class not a member of itself’ have no application.
· I haven’t examined any the other of the apparently related problems – I don’t know if they can be resolved in the same way or not
The rest below is just an idea.
Of course my rule about nouns and verbs is still need to block the f(f) iff ~f(f) paradox . Consider the f such that ~f(f).
f is a verb so cannot be used as a noun. I’ve prohibited that. There is a F such that symbol1(S,F,f,t). Is there an f such that ~f(F) where symbol1_r(S,F,f,t)?
There is no paradox here as ~f(F) doesn’t imply f(F).
Similarly with the relation T(R,S) defined as ~R(R,S) ; R and S are verbs, so cannot appear as nouns.
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