David Brightly comments:

. . . my old copy of Alan Hamilton,

Logic for Mathematicians, CUP 1978, uses 'statement variables' in his account of the 'statement calculus', as here. The justification for 'variable' is surely that statements have values, namely truth and falsehood. The truth value of a compound statement is calculated from the truth values of its component simple statements by composition of the truth functions corresponding to the logical connectives. This is analogous to the evaluation of an arithmetic expression by composition of arithmetic functions applied to the values of arithmetic variables.

I detect a possible conflation of two senses of 'value.' There is 'value' in the sense of truth value, and there is 'value' in the sense of the value of a variable.

If I am not mistaken, talk of truth values in the strict sense of this phrase enters the history of logic first with Gottlob Frege (1848-1925). Truth and Falsity for him are not properties of propositions, but values of propositional functions. Thus the propositional function denoted by 'x is wise' has True for its value with Socrates as argument, and False for its value with Nero as argument. Please note the ambiguity of 'argument.' We are now engaging in MathSpeak. The analogy with mathematics is obvious. The squaring function has 4 for its value with 2 or -2 as arguments. Propositional functions map their arguments onto the two truth values.

But we also speak in a different sense of the value of a variable. The bound variables in

(x)(x is a man --> x is mortal)

range over real items. These items are the values of the bound variables but they are not truth values. Therefore, one should not confuse 'value' in the sense of truth value with 'value' in the sense of value of a variable. When Quine famously stated that "To be is to be the value of a [bound] variable" he was not referring to truth values.

Brightly says that "The justification for 'variable' is surely that statements have values, namely truth and falsehood." I think that is a mistake that trades on the confusion just exposed. Agreed, statements have truth values. But it doesn't follow that that placeholders for statements are variables.

I was pleased to see that Hamilton observes the distinction I drew several times between an abbreviation and a placeholder. He uses 'label' for 'abbreviation,' but no matter. But I distinguish a placeholder from a variable while Hamilton doesn't.

To appreciate the distinction, first note that with respect to variables we ought to make a three-way distinction among the variable, say 'x,' the value, say Socrates, and the substituend, say 'Socrates.' Now consider the argument:

Tom is tall or Tom is fat

Tom is not tall

-------

Tom is fat

This argument has the form of the Disjunctive Syllogism:

P v Q

~P

-------

Q.

Obviously, 'P' and 'Q' are not abbreviations (labels); if they were then the second display would not display an argument *form*. It would be an abbreviated *argument*. But it doesn't follow that 'P' and 'Q' are variables. For if they were variables, then they would have both substituends andf values. But while they have substituends, e.g., the sentences 'Tom is tall' and 'Tom is fat,' they don't have values. Why not? Because we are not quantifying over propositions (or statements if you prefer). There are no quantifiers in the form diagram. (This is not to say that one cannot quantify over propositions.)

'Tom' is tall' is one of many possible substituends for 'P.' But 'Tom is tall' is not the value of 'P.' For we are not quantifying over sentences. We are not quantifying over propositions either. So *Tom is tall* is also not a value of 'P.'

My thesis is that placeholders in the propositional calculus are arbitrary propositional *constants*. Since they are constants, they are not variables. It is a subtle distinction, I'll grant you that, but it seems necessary if we are to think precisely about these matters. But then one man's necessary distinction is another man's hair-splitting.

You also argue that London must wrongly decide that 'if roses are red then roses are red' (RR) is a contingency, because we say it can be seen as having the form 'P-->Q' and

in generalstatements of this form are contingencies. Indeed they are. But we don't so decide. We say this is a special case in which P and Q stand for the same simple sentence, 'roses are red', not different ones. P and Q are therefore either both true or both false and either way the truth function for --> returns true. Hence this special case is tautologous. We disagree that the move from RR to 'P-->Q' must be seen as an abstraction. We retain the information that P and Q stand for specific substatements within RR, which may themselves have internal structure. 'Form' is a device for making such structure explicit.

So you are saying that 'P --> Q' has a special case that is tautologous. But that makes no sense to me if RR has both forms. A sentence (understood to have one definite meaning) is tautologous if its logical form is tautologous, and if RR has the form 'P--> Q' then it it is not tautologous as an instance of that form. So you seem committed to saying that RR is both tautologous and not tautologous.

Isn't that obvious? If one and same sentence (understood to have one definite meaning) has two logical forms, one tautologous and the other non-tautologous, then one and the same sentence is both tautologous and non-tautologous -- which is a contradiction.

One solution, as I have suggested several times already, is to say that, while 'P --> P' is a special case of 'P -->Q,' namely the case in which P = Q, the two forms are not both forms of 'If roses are red, then roses are red.' Only one of them is, the first one. The second is a form of the first form, not a form of the English sentence.

Putting the problem as an aporetic hexad:

1. 'P -->P' is a special case of 'P --> Q'

2. If a proposition s instantiates form F, and F is a special case of form G, then s instantiates G.

3. 'P --> P' is a tautologous form.

4. 'P --> Q' is a non-tautologous form.

5. No one proposition instantiates both a tautologous and a non-tautologous form.

6. 'If roses are red, then roses are red' instantiates the form 'P --> P.'

The hexad is inconsistent. Phoenix and London agree on (1), (3), (4), and (6). The Phoenician solution is to reject (2). The Londonian solution is reject (5).

But the Phoenicians have an argument for (5):

7. The logical form of a proposition is not an accidental feature of it but determines the very identity of the proposition.

Ergo

8. If s instantiates form F, then necessarily, s instantiates F.

ergo

5. No one proposition instantiates both a tautologous and a non-tautologous form.

I've looked through the available literature and it is very hard to find precise statements or discussion of the notion of a schema or placeholder. Quine says "Schemata are logical diagrams of statements; the letters 'p', 'q' etc., by supplanting the component clauses of a statement, serve to blot out all the internal matter which is not germane to the broad outward structures with which our logical study is concerned". Not hugely helpful.

Bill, I think you said (correct me if I am wrong) that a placeholder or propositional schema is itself a statement or sentence. Could we then compare it to a statement (or function of statements) in some foreign language whose meaning is unknown to us?

But if we did, we have the apparent problem you have already mentioned. 'P' is a sentence of a foreign language. So it could mean something like 'Socrates is sitting or Socrates is not sitting', which is necessarily true, or 'Socrates is sitting and Socrates is not sitting' which is necessarily false, or just 'Socrates is sitting' which is not necessary. How can that be, according to you? How can the same placeholder represent both a contingent and non-contingent statement?

Apologies if I misunderstood you, I am writing in haste, bus to catch etc.

Posted by: Ed | Tuesday, May 13, 2014 at 11:32 PM

Aha. Quine again: "A truth function of letters 'p' and 'q', etc., is strictly speaking not a statement, of course, since the letters are themselves not actual statements but mere dummies in place of which

any desired statementsmay be imagined".From this it clearly follows that 'tom is tall implies tom is tall' instantiates 'P implies Q'. For according to Quine, we may substitute any desired statement - tom is tall - for 'P', and any desired statement - tom is tall - for 'Q'. This gives 'tom is tall implies tom is tall'.

Posted by: Ed | Tuesday, May 13, 2014 at 11:45 PM

Do you have the reference for the Quine quotation?

I didn't say that a placeholder or propositional schema is itself a statement or sentence.

'All S are P' is a prop. schema but not a proposition or statement or sentence. This assumes of course that 'S' and 'P' are not mere abbreviations.

I agree with both quotations from Quine, although we now have a another word to fret over, 'dummy.'

What you say clearly follows does not so clearly follow. It depends on what exactly Quine meant in the second quotation. Obviously, all of the following are instances of the schema 'P --> Q':

If Tom is tall, then Tom is fat

If Sally is hot, then Fiona is cooking dinner

If Trisha is swimming and Mary is biking, then Tom is reading Quine

and so on.

But it is not so obvious that 'If Tom is tall, then Tom is tall' --assuming no equivocation, nota bene!-- is a substitution instance of the schema in question.

If you say it is, then we get the inconsistent hexad above, which it would be nice if you commented on.

Posted by: BV | Wednesday, May 14, 2014 at 05:42 AM

>>The Londonian solution is reject (5) [No one proposition instantiates both a tautologous and a non-tautologous form].

Not London, rather, the standard/logical/textbook solution. "To be a tautology, then, a sentence must be derivable by substitution from at least one tautologous form, but it must not be thourght that tautologies are derivable only from tautologous forms. For while 'A v ~A' is derivable from the tautologous form 'p v ~p', it is also derivable from the non-tautologous 'p v q' (substituting 'A' for 'p' and '~A' for 'q'). Whereas tautologous form yield only tautologies upon substitution, non-tautologous forms can yield both tautologies and non-tautologies". (

Elementary Symbolic Logic, William Gustason, Dolph E. Ulrich, p.43).I suppose you will say this is argument from authority. But there are not a few cases where only argument from authority is appropriate. Example: the elementary chemical formula for water, the standard American spelling for the English 'colour', the first letter of the alphabet. You can find the answer to these in any dictionary or standard textbook. It is not appropriate to object 'argument from authority' in such cases.

Posted by: london ed | Wednesday, May 14, 2014 at 05:42 AM

I see our comments crossed in the aether. I'll be back later. Now I must attend to the needs of Bro Jackass, Fratre Asino, in the Italian of St. Francis.

Posted by: BV | Wednesday, May 14, 2014 at 05:47 AM

Hello Bill, and thank you for the lengthy reply.

First of all, can we agree that neither of us is confusing placeholders/propositional variables with the variables of the predicate calculus? One way of making this clear is to point out that the

languageof the predicate calculus has syntactic elements, often the letters 'x', 'y',..., but in more formal treatments the letter 'x' with suffices, which appear in quantifications. These are variables within the language of the predicate calculus. On the other hand, the propositional calculus has syntactic elements 'p', 'q', 'r',..., or more formally the letter 'p' with suffices, as in Hamilton, which stand, as you say, for 'arbitrary propositional constants'. There is no sense in which, in a single wff of the prop calc, such as 'p-->q', p and q 'range over' some set of values, whereas in the '∀x.Ax' of the pred calc there is a sense that x ranges over some set of values or objects. We agree so far I hope? Nevertheless, there is a clear sense, when we talkaboutthe prop calc, rather than merely expressing ourselvesinit, that the ps and qs can stand for any statements at all. That is the sense of your 'arbitrary' that I quoted above. Perhaps we can encapsulate this by saying that the ps and qs are variables in themeta-theoryof the prop calc.Regarding the characterisation of RR as a tautology, etc, we would have to say this: a sentence is a tautology iff it instantiates a tautologous form; a sentence is a contradiction iff it instantiates a contradictory form; a sentence that is neither a tautology nor a contradiction is a contingency. RR instantiates the tautologous form 'P-->P' and hence is a tautology. Note the similarity with the London treatment of validity of arguments.

I can appreciate that, from your point of view regarding form, it makes no sense to say that RR has both a tautologous and a contingent form, so you reject (2). I'm not claiming that your view is untenable, though I agree with Ed and Colin Cheyne that it makes for a messy metatheory. And isn't a theory of the form of form going to be bedevilled by an analogous issue, one level of abstraction up, as it were? However, I am concerned to rebut your claims that

wehave got it wrong. To that end I reject your (7). Your notion of form is analogous to 'species' in the sense of 'most specific genus', I think. We see 'form' as analogous to 'genus'. But I don't see how its species 'determines the very identity' of a thing, any more than a genus does.Posted by: David Brightly | Wednesday, May 14, 2014 at 06:06 AM

>>But it is not so obvious that 'If Tom is tall, then Tom is tall'

I was careful to emphasise '

anydesired statement', and refer you again to what Quine says. (MoL, p.22, although my edition differs from modern ones).You understand the meaning of the word 'any', yes?

Posted by: london ed | Wednesday, May 14, 2014 at 06:06 AM

Well you could, but honestly mate I think you are on a hiding to nothing. It's like you with a knight fighting against London with a rook and a pawn not far from queening, and you making all sorts of valiant and bravely intended subtle moves against the impending and inevitable doom. I admire your pluck but, really, give that king a little push into oblivion and we are done and have a pint or two at the bar.

Posted by: london ed | Wednesday, May 14, 2014 at 06:13 AM

Aha again. Quine,

Elementary Logicp.37, says thatJones is ill & ~(Jones is ill & Smith is away)

is a substitution instance of both 'p & ~(p & q)' and 'p & ~(q & r)', and he explicitly adds, and I quote "

there is no rule against substituting the same statement 'jones is ill' thus for different letter 'p' and 'q'", which is the London rule (or rather the London permission).Pawn has just queened, and mate in two.

Posted by: london ed | Wednesday, May 14, 2014 at 06:37 AM

Ed,

I of course accept the authority of Quine, and your appeal to it is justified when it comes to establishing what the official or mainstream logical doctrines are. You have now conclusively established what the official doctrine is when it comes to substitution. But what the doctrine is and whether it is philosophically sound are two different questions. I have been raising the second question. Your appeal to authority becomes fallacious if you think that Quine's pronouncement conclusively shows what the truth is on the matter under discussion.

But I have two more moves as I try to mate you with my bare knight (which is possible by the way!)

First, there is the difference that Brightly is sensitive to, namely, the difference between instantiating a sentence form and translating a sentence. I fully understand why you say that 'If Tom is tall, then Tom is tall' is a substitution-instance of 'P --> Q.' The justification for the London permission in this instance is that one can abstract away from the plain fact that the conditional sentence involves exactly one proposition.

But if we move in the opposite direction, from the sentence to the form, then it seems clear to me that 'If Tom is tall, then Tom is tall' does NOT have the form 'P --> Q.' For this form allows, though it does not require, the consequent to be a different proposition from the antecedent, and this in violation of what we clearly KNOW, namely, that exactly one proposition is involved.

So there is a problem here and that is what I am focusing on.

The second point is that you may be committing an ignoratio elenchi against me. We are not disputing what authoritative texts say. We are disputing whether the official doctrine is in the clear, i.e., is unproblematic.

Posted by: BV | Wednesday, May 14, 2014 at 11:28 AM

David,

I agree with your first paragraph, except that I would say that there are placeholders in both the prop calc and the pred calc. For example, if we symbolize 'Al is fat' as 'Fa,' 'a' is a placeholder, an arbitrary individual constant, and 'F' is a placeholder, an arbitrary predicate constant. Neither are variables. 'Fx' features a free individual variable and an arbitrary predicate constant. In second-order logic one can introduce symbols that are predicate variables. We would then be quantifying over properties.

Suppose we want to say that every first-order property is instantiated. I am font-challenged, but this will give the idea:

(P)(Ex)Px For every first-order property P, there is an x such that x instantiates P.

'(Ex)(P) Px' on the other hand would say that something has every property.

I hope to respond to the rest of your comment later.

Posted by: BV | Wednesday, May 14, 2014 at 11:44 AM

Well I'm off for a pint of English style milk stout. I shall ponder these subtleties for now. Good night all.

Posted by: Ed | Wednesday, May 14, 2014 at 12:00 PM

Bill,

Consider the following parody:

A Square is an equi-sided Rectangle; a Rectangle is an equi-angular Parallelogram; a Parallelogram is a Trapezium with two pairs of parallel sides; a Trapezium is a Quadrilateral with a pair of parallel sides. Let the forms be Square, Rectangle, Parallelogram, Trapezium, Quadrilateral, and let the shapes be instances of the forms. Then,

1. Rectangle is a special case of Parallelogram.

2. If a shape s instantiates form F, and F is a special case of form G, then s instantiates G.

3. Rectangle is an equi-diagonal form.

4. Parallelogram is a non-equi-diagonal form.

5. No shape instantiates both an equi-diagonal form and a non-equi-diagonal form.

6. The shape ☐ instantiates the form Rectangle.

is an inconsistent hexad. We can resolve the inconsistency by rejecting (2) or (5). Let's reject (2). For we can justify (5) with

7. The form of a shape is not an accidental feature of it but determines the very identity of the shape.

Ergo

8. If s instantiates form F, then necessarily, s instantiates F.

Ergo

5. No one shape instantiates both a equi-diagonal and a non-equi-diagonal form.

Posted by: David Brightly | Wednesday, May 14, 2014 at 03:49 PM

Ed, Where on earth can you get a pint of milk stout these days!? Draught Mackeson? I'm sure it will look good, taste good and, by golly, do you good.

Posted by: David Brightly | Wednesday, May 14, 2014 at 03:55 PM