5
$\begingroup$

At the end of ✳96.48, $ \sim(w=\overset{\smile}{R}‘max_R‘J_R‘x)$ is chosen over $ w\neq\overset{\smile}{R}‘max_R‘J_R‘x$, on account of the latter's implication of existence. But ✳13.02 states that they are the same.

✳13.02 $ x\neq y .=.\sim(x=y) $ Df

I wonder what I have missed. Where does PM say identity has anything to do with existence? Thanks,

enter image description here

$\endgroup$
1
  • $\begingroup$ Thanks, @Mauro. Same "sets of consequences" is exactly what I wanted to say but didn't figure out how to put it. $\endgroup$ May 10 '14 at 18:33
4
$\begingroup$

After a sleepless week I've decide to ask to Prof.Gregory Landini (Iowa University), my preferred "interpreter" of Russell's logic.

Here is his kind answer; I'll give it verbatim :

This is a nice passage that illustrates something I've been arguing for years -- viz., definintions formed with individual variables cannot be applied to definite descriptions (because definite descriptions are not genuine terms). I didn't know it and find is wholly enjoyable that you found it and brought it to my attention.

According to ✳30.01 $R‘y = (\iota x)(xRy)$ Df; thus, in our formula, "$\overset{\smile}{R}‘ ...$" is a definite description of a class and not a class term. Thus there is an issue of scope of the description.

In order to apply the definition ✳13.02 [i.e. $x≠y.=.∼(x=y)$ Df] one must first eliminate the definite description. Note as well that when scope marker are dropped, smallest (most secondary scope possible) is intended.

The general form of the case at hand is this:

$w \ne \iota aFa =_{df} [\iota aFa]( w \ne \iota aFa)$.

This is the smallest scope possible -- because definition ✳13.02 is framed with individual variables and thus cannot apply until after the definite description has been eliminated.

In contrast,

$\lnot(w = \iota aFa) =_{df} \lnot [\iota aFa](w = \iota aFa)$

The two scopes are not equivalent.

Added

This is explained by the "paradigmatic" case regarding "the King of France" and the management of scope in definite descriptions.

See Principia page 70 :

$\quad \quad \quad \quad \quad [(\iota x) (\phi x)]. \lnot \psi(\iota x)(\phi x)$

will mean $\quad \quad (\exists c): \phi x . \equiv_x. x = c: \lnot \psi c$,

while $\quad \quad \lnot \{ [(\iota x) (\phi x)]. \psi(\iota x)(\phi x) \}$

will mean $\quad \quad \lnot \{ (\exists c): \phi x . \equiv_x. x = c: \psi c \}$.

Here again, when $(\iota x) (\phi x)$ does not exist, the first is false and the second true.

Thus, in ✳96.48, $\lnot(w=\overset{\smile}{R}‘ ...)$ corresponds to the second case, in which $\overset{\smile}{R}‘...$ does not exists.

$\endgroup$
3
  • $\begingroup$ Absolutely insightful! By *30.01, $\overset{\smile}{R}‘max_R‘J_R‘x$ is an incomplete symbol, therefore not an individual. Yes, there is an issue of scope. It is discussed in the summary of *14. So many thoughts are racing through my head now. I need to study some more. Thanks you so much, @Mauro, for your help. And please pass my thanks to Prof. Gregory Landini also. $\endgroup$ May 18 '14 at 18:44
  • $\begingroup$ @GeorgeChen - I've made some addition. You are right : it's not easy to find a fault into W&R's Principia ... $\endgroup$ May 18 '14 at 19:46
  • $\begingroup$ Thanks, @Mauro, for bringing my attention back to the King's chapter, every line of which is now full of meanings. Yes, that is where the ultimate answer is. $\endgroup$ May 19 '14 at 18:03
2
$\begingroup$

W&R's comment above is not clear to me.

According to logical rules, an abbreviation is only a "symbol": it cannot alter the theorems provable in the system.

Thus, if $x \ne y$ is (as usual) an abbreviation for $\lnot (x=y)$, we have no possibility of having different "sets of consequences".

I think that the comment is related to ✳13.19 : $\vdash ∃y(y=x)$, which is a correct logic rule, and to the "missing" semantics of PM.

In "modern" first-order logic, the law : $\vdash ∃y(y=x)$ is (universally) valid because we assume that every universe of discourse (i.e.every domain of interpretation) is not empty. Thus, in every interpretation, we have at least one object, which is for sure "equal to itself".

The above law is usually derived from the identity axiom : $x=x$ through the $\exists$-introduction rule :

from $\varphi(t)$, infer $\exists x \varphi(x)$.

If we apply the rule to $y \ne x$ (which is exactly : $\lnot (y=x)$), we can infer $∃y (y \ne x)$, but now we do not obtain a logical law.

The formula $∃y (y \ne x)$ is not (universally) valid because it is false in every domain with at least two objects, exactly like $y \ne x$.

I think that in ✳96.48 the free variable $w$ into $\sim(w=\overset{\smile}{R}‘max_R‘J_R‘x)$ must be read as implicitly universally quantified.

If so, the sub-formula is equivalent to $∀w \sim(w=\overset{\smile}{R}‘max_R‘J_R‘x)$ i.e.to $\sim \exists w (w=\overset{\smile}{R}‘max_R‘J_R‘x)$.

A possible reading of the annotation can be this: W&R want to avoid the misconception related to the above rule.

If we apply it to the formula $w\neq\overset{\smile}{R}‘max_R‘J_R‘x$ we can infer $\exists w (w\neq\overset{\smile}{R}‘max_R‘J_R‘x)$, which is not $\sim \exists w (w=\overset{\smile}{R}‘max_R‘J_R‘x)$.

$\endgroup$
3
  • $\begingroup$ Thanks, @Mauro. Give me a few more days. It seems that I can never spend enough time on introduction and part one. $\endgroup$ May 11 '14 at 11:08
  • $\begingroup$ I messed up the tail and the circle in the last comment.I agree that ∃y(y≠x) can't be taken for granted. w always exists. $\overset{⌣}{R}‘max_R‘J_R‘x$ exists only if $\overset{←}{R}_✳‘x$(the posterity of x with respect to R) is shaped like a Q. If $\overset{←}{R}_✳‘x$ consists of only $I_R‘x$(the circle) or only $J_R‘x$(the tail), then there is no junction to speak of and $\overset{⌣}{R}‘max_R‘J_R‘x$ does not exist. Still I agree with you that an abbreviation makes no difference in consequences. $\endgroup$ May 12 '14 at 15:27
  • $\begingroup$ A possible workaround is to interpret $w=\overset{\smile}{R}‘max_R‘J_R‘x$ in terms of *30.3, i.e. $w$ is the one that has relation $\overset{\smile}{R}$ to $max_R‘J_R‘x$. Also, PM 1st ed is not free from harmless mechanical errors, as is shown by *93.132 which was corrected in 2nd ed, but I'm very reluctant to admit faults in PM. $\endgroup$ May 15 '14 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.