4
$\begingroup$

The first edition of Principia Mathematica clearly distinguishes "Socrates is a man" and "'Socrates is a man' is true." Judging from the context, the distinction is neither a primitive idea nor a definition; it is something the reader is supposed to "see." For some reason I can't "see" it, although I can deliberately memorize the distinction the same way I memorize multiplication table. I have an inkling that this distinction should directly appeal to the sense, like the distinction between red and blue. If you can "see" this distinction, please kindly help me out.

The following passage is from page 41, Introduction, Chapter II, section II, The Nature of Propositional Functions. enter image description here http://i.stack.imgur.com/ejGD1.png

These symbols are for your convenience:

propositional function: $\phi(\hat{z})$

Ambiguous value: $ \phi(z) $

A function with itself as an argument: $\phi(\phi(\hat{z}))$

$\endgroup$
  • 1
    $\begingroup$ It seems to say "quadruplicity drinks procrastination" is nonsensical; but "'quatruplicity drinks procrastination' is true" is false, and therefore, not nonsense. $\endgroup$ – George Chen Jan 18 '14 at 23:37
  • 1
    $\begingroup$ You say a distinction is being made. In his book Introduction to Mathematical Philosophy he makes no such distinction, in fact, he claims they mean the same. As for Principia Mathematica, in its introduction he says "Thus when a text-book of logic asserts that “A is A,” without any indication as to what A may be, what is meant is that any statement of the form “A is A” is true", supporting the non-distinction you claim to exist. Can you please pinpoint where such a distinction is made? $\endgroup$ – Git Gud Jan 18 '14 at 23:45
  • $\begingroup$ Related. $\endgroup$ – Git Gud Jan 18 '14 at 23:47
  • $\begingroup$ It's in chapter II of introduction. At the beginning of this section, it says "the values of the function are propositions," from which I infer the authors wanted to say the values of a propositional function are not truth-values. And in his An Inquiry into Meaning and Truth (First published in 1940), there are similar distinctions, from which I infer Russell didn't change his mind regarding this point. $\endgroup$ – George Chen Jan 19 '14 at 0:45
2
$\begingroup$

A possible interpretation of the passage must take into account the context, that is : Chapter II : THE THEORY OF LOGICAL TYPES.

In particular, the passage is taken from section II. The Nature of Propositional Functions.

We must remember the lack of clear and systematic distinction in W&R's Principia between object-language and meta-language (and meta-theory); in this chapter, they are explaining the "syntactical" restrictions involved in type theory.

See page 40 :

Now given a function $ \phi(\hat{x})$, the values for the function are all propositions of the form $\phi x$. It follows that there must be no propositions, of the form $\phi x$, in which $x$ has a value which involves $\phi \hat x$.

That is to say, the symbol " $\phi (\phi \hat x$) " must not express a proposition, as " $\phi a$ " does if $\phi a$ is a value for $\phi \hat x$. In fact " $\phi (\phi \hat x$) " must be a symbol which does not express anything: we may therefore say that it is not significant.

Roughly, the formula $\phi(\phi)$ is not true nor false; it is simply meaningless, because it violates what we today will call the "rules of formation"; see page 41 :

Since "$(x) . \phi x$ " involves the function $\phi \hat x$, it must, according to our principle, be impossible as an argument to $\phi$. That is to say, the symbol " $ \phi \{(x) . \phi x \} $ " must be meaningless.

Gregory Landini, in Russell's Hidden Substitutional Theory (1998), page 279, proposes a reading that starts from Frege's hierarchy of "levels" of functions. According to Landini :

The idea is that a "function" (i.e., a predicate variable) can occur in a subject position (argument position) of another predicate variable only if this position represents a predicate position in the semantics [Landini points to pages 47 and 48 of PM].

$\endgroup$
  • $\begingroup$ Thanks, @Mauro. "$\phi(a)$ is true" looks cumbersome, but according to this book page 179, most of educated knowledge(as opposed to personal experiences) is actually of this type. Taking this as a hint, I think the first one is supposed to invoke a mental image; the second one is merely or mostly verbal. $\endgroup$ – George Chen Jan 19 '14 at 22:17
  • $\begingroup$ By PM's standards, "$\phi(x)$ is true" is one order above $\phi(x)$. That's the technical difference. $\endgroup$ – George Chen Feb 11 '16 at 18:01
2
$\begingroup$

The only thing I can think of is that "Socrates is a man" is a sentence in the sense that it has a truth value, but we do not make any claim as to what that truth value may be. The claim that "Socrates is a man" is true, is asserting not only that the same sentence has a truth value, but also that that truth value is True.

Just to round things off: "Socrates slept quickly" is not even a sentence since it has no truth value at all, it is simply nonsense written in a grammatically plausible fashion. However, "Socrates had exactly 100,000 hairs on his head by his 18th birthday" is a sentence. It may be true or false, but we'll never know.

I hope this is at all in the right direction to answer your question.

$\endgroup$
  • 1
    $\begingroup$ Thanks @Ittay. That helps. Look like I need to improve my understanding of what an assertion is in this context. $\endgroup$ – George Chen Jan 19 '14 at 0:05
  • $\begingroup$ @GeorgeChen, please use Mathjax if you can, and ASCII if you can't. Those symbols don't work for everyone. $\endgroup$ – dfeuer Jan 19 '14 at 9:00
  • $\begingroup$ @dfeuer, Wilco. $\endgroup$ – George Chen Jan 19 '14 at 9:53
  • $\begingroup$ Can I say "'$\phi(a)$ is true' is false" implies "'$\phi(a)$ is false' or '$\phi(a)$ is nonsense'?" $\endgroup$ – George Chen Jan 19 '14 at 10:15
  • $\begingroup$ @George Chen - I think we must pay attention to not mixing : (i) the interpretation of W&R's PM, with (ii) modern usage in mathematical logic. About (i), the issues regarding the correct interpretation of PM's "philosophical" parts, like the motivation behind type-theory, are complex. About (ii), the distinction between language and meta-language was at the core of Tarski's truth definition and Ramsey's approach to "semantic" paradoxes (like "the Liar"). $\endgroup$ – Mauro ALLEGRANZA Jan 20 '14 at 11:06
2
$\begingroup$

I have never read or otherwise studied the Principia; however, I think the general distinction to which Russell is alluding is still very much a recognized principle in modern (formalized) mathematics. Its basically the difference between a sentence $\varphi,$ versus the metasentence $\vdash \varphi$.

Conceptually, the distinction is best explained with reference to partially ordered sets (hereafter poset). In a poset, we can assert $x \leq y$ (intuitively, $x$ entails $y$). We may also have a meet-semilattice structure, in which case our assertions can be more sophisticated: we may write $x \wedge x' \leq y,$ intuitively asserting that $x$ and $x',$ taken together, entail $y$.

Note that $\wedge$ is a function, $\leq$ a relation.

Now furthermore, any given meet-semilattice may or may not admit the existence of a function $\rightarrow$ with the following property.

  • $x \wedge x' \leq y$ iff $x \leq x' \rightarrow y$.

If such a function exists, it is unique, by this result. (If it is not clear what the above definition has to do with Galois connections, please comment and I will clarify.)

Anyway, if there is such a function (which I will call "implication"), then it can be added to the language (alongside $\wedge$ and $\leq$) to get a more expressive language. And we can prove the basic facts we expect from implication, such as modus ponens:

$$(x \rightarrow y) \wedge x \leq y$$

By the way, I recommend saying $\leq$ as "entails", and $\rightarrow$ as "implies", although this is not standardized.

Anyway, the point is that $\rightarrow$ can be conceived as an internalization of $\leq.$ Note that $\rightarrow$ is a function, while $\leq$ is a relation. Thus, spiritually, we can think of $x \rightarrow y$ as a statement internal to the language, while a formula like $x \leq y$ can (kind of) be viewed as part of the metalanguage. I am speaking very informally, here, of course.

Now I haven't really explained how $\rightarrow$ is an internalization of $\leq$, so lets do that. It turns out that if a function $\rightarrow$ with the property of interest exists, then so too does a top element, so long as we're working in a non-empty poset. (Hint: consider the expression $x \rightarrow x$). Denote the top element $\top;$ we can think of this as denoting unadulterated truthood. Furthermore, it can be shown that $x \leq y$ is equivalent to $\top \leq (x \rightarrow y)$. This is the sense in which $\rightarrow$ is an internalization of $\leq$.

Finally, lets switch to more logical notation. Instead of $\leq$, write $\vdash$ (this can also be articulated: "entails"). And lets move to greek letters, which can be thought of as denoting logical formulae. Furthermore, as shorthand for $\top \vdash \varphi$, let us write $\vdash \varphi.$

Then there is a clear difference between $\varphi$, and $\vdash \varphi$.

However, oftentimes $\varphi$ can be used as shorthand for $\vdash \varphi$, if the meaning is clear in context. Similarly, sometimes $\varphi \rightarrow \psi$ can be used as shorthand for $\vdash \varphi \rightarrow \psi$, or in other words $\varphi \vdash \psi$.

I think this is (at least conceptually) the distinction to which Russell is alluding.

$\endgroup$
  • $\begingroup$ I appreciate your perspective. $\endgroup$ – George Chen Jan 19 '14 at 10:23
  • $\begingroup$ @GeorgeChen, you're welcome. Feel free to comment with any questions you may have. $\endgroup$ – goblin Jan 19 '14 at 10:34
  • 2
    $\begingroup$ @GeorgeChen - user18921's answer : "Its basically the difference between a sentence $\phi$, versus the metasentence $\vdash \phi$" point (for me) in the right direction. I think that the historically perspective can be useful. The symbol "$\vdash$" dates back to Frege, in Begriffsschrift (1879) [pag.11 of van Heijenoort edition] : "A judgement will always be expressed by means of the sign $\vdash$ [...]. If we omit the small vertical stroke at the left end of the horizontal one, the judgement will be transformed into a mere combination of ideas [Vorstellungsverbindung] ..." 1/3 $\endgroup$ – Mauro ALLEGRANZA Jan 19 '14 at 16:43
  • 1
    $\begingroup$ "... of which the writer does not state whether he acknowledges it to be true or not. For example, let $\vdash A$ stand for the judgment "Opposite magnetic poles attract each other"; then $\textminus A$ will not express this judgment; [...] When the vertical stroke is omitted, we express ourselves paraphrastically, using the words "the circumstance that" or "the proposition that". Not every content becomes a judgment when $\vdash$ is written before its sign; for example, the "house" does not. We therefore distinguish contents that can become a judgment form those that cannot." 2/3 $\endgroup$ – Mauro ALLEGRANZA Jan 19 '14 at 16:50
  • 1
    $\begingroup$ So, I think, historical context can help us : W&R are pointing to the distiction between using a symbol for expressing "a content" and the assertion that that content "is a fact", i.e.making a judgment. Modern usage of $\vdash$ to mean "is a theorem" dtaes back to Rosser and Kleene (1934-35) [see Kleene, Mathematical Logic (1967), pag.36, footnote 30]. The link, I suppose, is this : you can have a formula $\phi$ of your language, and it can be true or false (or neither, if it not closed). When you assert $\vdash \phi$ you are asserting (in the metalanguage) that you have a proof of it. $\endgroup$ – Mauro ALLEGRANZA Jan 19 '14 at 17:05
0
$\begingroup$

According to the Theory of Types, "Clifford the Red Dog is not a colour" is nonsense, but "'Clifford the Red Dog is a colour' is not true" is meaningful.

Let $a$ be an individual, $\beta$ be a class of individuals, $\kappa$ a class of classes, $a \in \beta$ is either true or false, $\beta \in \kappa $ is either true or false, but $a \in \kappa$ is nonsense. You can't say $a \notin \kappa$, but you can say "$a \in \kappa$ is not true."

"$\sin(0) = 0$" is true; "$\sin(0) = 1 $" is false; but "$\sin(Socrates) = 0.5 $" is neither true nor false, it is nonsense. "$a \notin \kappa$" is as meaningless as "$sin(Socrates) = 0.5$".

When $\phi(x)$ is not true, $\phi(x)$ can be either false or nonsense.

A sentence that is significant but fails to form a corresponding mental state due to the reader's lack of experience is merely verbal and is nonsense to such a reader. The following is an excerpt from Huxley's Brave New World. It demonstrates the difference between "$\phi(x)$" and "$\phi(x)$ is true."

At breakfast the next morning, "Tommy," some one says, "do you know which is the longest river in Africa?" A shaking of the head. "But don't you remember something that begins: The Nile is the ..."

"The - Nile - is - the - longest - river - in - Africa - and - the - second - in - length - of - all - the - rivers - of - the - globe ..." The words come rushing out. "Although - falling - short - of ..."

"Well now, which is the longest river in Africa?"

The eyes are blank. "I don't know."

"But the Nile, Tommy."

"The - Nile - is - the - longest - river - in - Africa - and - second ..."

"Then which river is the longest, Tommy?"

Tommy burst into tears. "I don't know," he howls.)

$\endgroup$
  • $\begingroup$ It took me two months of brooding to make this connection. $\endgroup$ – George Chen Mar 18 '14 at 17:01

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.