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Methodology and ontology..Beyond the 'end of days'

Post 221

Bx4

Hi psi

Sorry for delay.

1. Got sucked into a series of pointless debate on one of the Mustarland Redux board.

2. Hid my IP address behind a proxy server and had a devil of a time logging on to h2g2.

3. Can't find Merricks paper online any more.

So this is something of a holding post:

>>I'm not clear why you think this is the chase<<

Because excluding all side issues our substantive discussion is that Merricks conclusion is circular or not. I argue that is because it is implicit in the definition:

(a)'S knows that p' if and only if S's belief is /warranted/ and p is true

whereas you contend that it is not.

IIRC, Merrick not only does not label the propositions on either side of the biconditional, make any reference to the biconditional expansion or its components. So our discussions of these are essentially side issues and not relevant to the main discussion. I am quite happy to discuss them separately but I want to concentrate on the nature of biconditional definitions first.

I introduced

(b) KsP<=>BsP.Wp.p

as the symbolic form of (a). I now think this is wrong since 'S believes P' is not present in the biconditional hence the RHS of the biconditional contains only /two/ propositions:

(1) S's belief is /warranted
(2) p is true

which, symbolically, gives something like

(c) Ksp<=>BsW.p

This seems a good place to until I know whether or not you agree to this revision.

bs


Methodology and ontology..Beyond the 'end of days'

Post 222

Psiomniac

Hi Bx4,

>>Because excluding all side issues our substantive discussion is that Merricks conclusion is circular or not. I argue that is because it is implicit in the definition:


(a)'S knows that p' if and only if S's belief is /warranted/ and p is true

whereas you contend that it is not. <<

I do agree that the issue of circularity is the chase. Indeed, I think I have not merely contended, but shown that it is not implicit in the definition.

>>IIRC, Merrick not only does not label the propositions on either side of the biconditional, make any reference to the biconditional expansion or its components. So our discussions of these are essentially side issues and not relevant to the main discussion. I am quite happy to discuss them separately but I want to concentrate on the nature of biconditional definitions first.<<

I disagree, since the definition and the biconditional are logically equivalent.

>>I introduced

(b) KsP<=>BsP.Wp.p

as the symbolic form of (a). I now think this is wrong since 'S believes P' is not present in the biconditional hence the RHS of the biconditional contains only /two/ propositions:

(1) S's belief is /warranted
(2) p is true

which, symbolically, gives something like

(c) Ksp<=>BsW.p <<

S's belief is warranted seems to me to be logically equivalent to (S believes p).(S's belief that p is warranted). If you think that's not the case, perhaps you could show how? Otherwise we might as well stick with what we have done rather than reinvent the wheel.

>>This seems a good place to until I know whether or not you agree to this revision. <<

I can't see how your revision would help your case, but maybe you have an argument in mind?

ttfn


Methodology and ontology..Beyond the 'end of days'

Post 223

Bx4

hi psi

>>I disagree, since the definition and the biconditional are logically equivalent.<<

I didn't say they weren't merely, iirc' that Merricks doesn't use the biconditional expansion but does use the unexpanded biconditional.

Moreover since the are 'logically' equivalent then I assume you would agree that the outcomes of their truth tables are identical so why use the more complex formula when the simpler will do.

>>S's belief is warranted seems to me to be logically equivalent to (S believes p).(S's belief that p is warranted). If you think that's not the case, perhaps you could show how? Otherwise we might as well stick with what we have done rather than reinvent the wheel. <<

Haven't I already done so when I pointed out that Merricks' biconditional does not contain the phrase 's believes that p'.

I have already made //my// argument as to why I consider Ksp<=>BsW.p are more accurate representation of Merricks' biconditional than Ksp<=>BsP.Wp.p which is simply that the proposition 's believes that p' does not appear in the biconditional.

So surely it is up to you to present argument in support of //your// claim that the two are logically equivalent rather than using the sophistical gambit of attempting to shift the burden of proof from you to me.

In any case, allowing, arguendo, that the two //are// logically equivalent then again, presumably, the outcomes of the truth tables of both are identical why would you prefer the more complex formula?

>>I can't see how your revision would help your case, but maybe you have an argument in mind?<<

Well, I may have and it may be that eliminating unneccessary complexity will help me make it. smiley - winkeye
bs


Methodology and ontology..Beyond the 'end of days'

Post 224

Psiomniac

Hi Bx4,

>>I didn't say they weren't merely, iirc' that Merricks doesn't use the biconditional expansion but does use the unexpanded biconditional.<<

But you also said:

>>So our discussions of these are essentially side issues and not relevant to the main discussion.<<

That's what I disagree with, since arguments that apply to your expanded form will also apply to Merricks' argument. In particular, my demonstration, using the expanded form, that the argument is not circular holds in virtue of the logical equivalence to the unexpanded form.

>>Moreover since the are 'logically' equivalent then I assume you would agree that the outcomes of their truth tables are identical so why use the more complex formula when the simpler will do. <<

Simply because I have already demonstrated the case using your more complex formulation. Since you accept the equivalence, it will not do for you to attempt to deflect the argument by saying Merricks' didn't use the expanded form; they are equivalent.

>>Haven't I already done so when I pointed out that Merricks' biconditional does not contain the phrase 's believes that p'. <<

No, see above. The equivalence is clear, the argument already made.

>>I have already made //my// argument as to why I consider Ksp<=>BsW.p are more accurate representation of Merricks' biconditional than Ksp<=>BsP.Wp.p which is simply that the proposition 's believes that p' does not appear in the biconditional.<<

Yours seems like a purely lexical argument rather than a logical one. In other words, Merricks uses particular phrases to specify propositions and you seem to think that the closer these are imitated the more accurate the logical representation, yet you offer no basis for this. This seems odd, given your knowledge of how many propositions it takes to rigorously translate English phrases (you pointed this out during the 'Onegin shot Lensky' discussion of fictional truth). I actually think that aping Merricks' English phrases when symbolizing produces a /less/ accurate formulation since in Ksp<=>BsW.p, the term BsW contains no mention of what S believes, namely p. So you have introduced ambiguity rather than accuracy.

>>So surely it is up to you to present argument in support of //your// claim that the two are logically equivalent rather than using the sophistical gambit of attempting to shift the burden of proof from you to me. <<

Do you agree that:

1) S's belief that p is warranted

entails that:

2) (S believes p).(S's belief that p is warranted) ?

The entailment is clear from the fact that in order for a S's belief that p be warranted, S must actually have that belief, otherwise there is referential failure? There is no attempt by me to shift any burden of proof here, there is no burden!

>>In any case, allowing, arguendo, that the two //are// logically equivalent then again, presumably, the outcomes of the truth tables of both are identical why would you prefer the more complex formula? <<

Because we have already done the work. Perhaps a more interesting question is, allowing, arguendo, that the two //are// logically equivalent and the outcomes of the truth tables of both are identical why would you prefer to go over it all again? The truth tables are clear, there is no circularity. What is to be gained by going through another, logically equivalent argument?

>>Well, I may have and it may be that eliminating unneccessary complexity will help me make it. <<

Feel free to go ahead and make it, although it seems to me you are on the horns of a dilemma- either they are logically equivalent, in which case simplifying will not help your case as it has already been defeated logically, or your simplified version is not logically equivalent. In that case, although you might attempt some legerdemain to show a spurious circularity, this will be based on syntactical resemblance to Merricks' phrases rather than logical considerations, which cannot be a sound basis upon which to argue for /logical/ circularity.

In my view your new formulation is ambiguous and less accurate and thus only logically equivalent if interpreted correctly.

Merricks' formulation is:

"S knows that p, therefore, if and only if S’s belief that p is warranted and p is true."

However your '(1) S's belief is /warranted ' omits any mention of p. So we have to interpret this as S's belief [that p] is warranted to retain logical equivalence.

ttfn


Methodology and ontology..Beyond the 'end of days'

Post 225

Bx4

Hi psi apologies. Got sucked in elsewhere.

hi psi

>>That's what I disagree with, since arguments that apply to your expanded form will also apply to Merricks' argument. In particular, my demonstration, using the expanded form, that the argument is not circular holds in virtue of the logical equivalence to the unexpanded form.<<

However since Merricks' doesn't use the expanded form and as both are equivalent then your argument can, presumably be made using the unexpanded form.

>>Simply because I have already demonstrated the case using your more complex formulation. Since you accept the equivalence, it will not do for you to attempt to deflect the argument by saying Merricks' didn't use the expanded form; they are equivalent.<<

However since I have already said that I don't need the expanded form to make my case and since it is unnecessarily complex I wonder why you are so set on adhering to it. Seems that it is you not me who is seeking to deflect the argument.

>>No, see above. The equivalence is clear, the argument already made.<<

Since Merricks' "S knows that p if and only if S's belief that p is warranted and P is true" does not contain the statement "S believes that p" then why would it appear in the expansion? Surely it is:

"if S knows that p then S's belief that p is warranted and P is true and if S's belief that p is warranted and P is true the S knows that p"

I begin to wonder if your penchant for overcomplexity is because you are unable to make your putative argument without it.


>>Yours seems like a purely lexical argument rather than a logical one. In other words, Merricks uses particular phrases to specify propositions and you seem to think that the closer these are imitated the more accurate the logical representation, yet you offer no basis for this. This seems odd, given your knowledge of how many propositions it takes to rigorously translate English phrases (you pointed this out during the 'Onegin shot Lensky' discussion of fictional truth). I actually think that aping Merricks' English phrases when symbolizing produces a /less/ accurate formulation since in Ksp<=>BsW.p, the term BsW contains no mention of what S believes, namely p. So you have introduced ambiguity rather than accuracy.<<


I have replaced one logical proposition with a simpler one, nothing more nothing less.

>>Do you agree that:

1) S's belief that p is warranted

entails that:

2) (S believes p).(S's belief that p is warranted) ?

The entailment is clear from the fact that in order for a S's belief that p be warranted, S must actually have that belief, otherwise there is referential failure? There is no attempt by me to shift any burden of proof here, there is no burden!<<

Well there is since you fail to consider the LHS of the biconditional; that is if you want to shoehorn 'S believes p' into the RHS of the biconditional then you have to also shoehorn it into the left, giving:

Bsp.KSp<=>BSP.BSp(W).p

and since BsP 'cancels out' nothing is served by including it. The burden of proof still lies with you!

>>Because we have already done the work. Perhaps a more interesting question is, allowing, arguendo, that the two //are// logically equivalent and the outcomes of the truth tables of both are identical why would you prefer to go over it all again? The truth tables are clear, there is no circularity. What is to be gained by going through another, logically equivalent argument?<<

Because the two propositions may be equivalent and the outcome of the two truth tables may be equivalent but the two truth tables are not equivalent in complexity and I am unclear why you are so averse to demonstrating the validity and soundness of your putative argument using the simpler proposition/truth table rather than devoting a whole post to coming up with less than wholly persuasive arguments to avoid doing so.

>>Feel free to go ahead and make it, although it seems to me you are on the horns of a dilemma- either they are logically equivalent, in which case simplifying will not help your case as it has already been defeated logically, or your simplified version is not logically equivalent. In that case, although you might attempt some legerdemain to show a spurious circularity, this will be based on syntactical resemblance to Merricks' phrases rather than logical considerations, which cannot be a sound basis upon which to argue for /logical/ circularity.<<

Methinks you protest too much.


>>
Merricks' formulation is:

"S knows that p, therefore, if and only if S’s belief that p is warranted and p is true."

However your '(1) S's belief is /warranted ' omits any mention of p. So we have to interpret this as S's belief [that p] is warranted to retain logical equivalence.<<

Simply resolved.

"S knows that p, therefore, if and only if S’s belief that p is warranted and p is true."

KSp<=>BSp(W).p

(posted in haste so usual apologies)

I was going to develop my argument here but, on reflection, better to do it separately.

bs


Methodology and ontology..Beyond the 'end of days'

Post 226

Psiomniac

Hi Bx4,

>>However since Merricks' doesn't use the expanded form and as both are equivalent then your argument can, presumably be made using the unexpanded form.<<

Yes presumably, but you have yet to offer a sensible reason for doing so.

>>However since I have already said that I don't need the expanded form to make my case and since it is unnecessarily complex I wonder why you are so set on adhering to it. Seems that it is you not me who is seeking to deflect the argument.<<

I adhere to it because I have already done the work. Then you say you have changed your mind and have changed your formulation to a simpler but logically equivalent one and you want me to go through the argument again, despite conceding that they are logically equivalent. When I ask why, you just reflect the question back and say it is me who is deflecting. Yet I have already made a perfectly good argument to which you have offered no counter. Instead you seem to exaggerate the complexity difference.

>>Since Merricks' "S knows that p if and only if S's belief that p is warranted and P is true" does not contain the statement "S believes that p" then why would it appear in the expansion? Surely it is:

"if S knows that p then S's belief that p is warranted and P is true and if S's belief that p is warranted and P is true the S knows that p"

I begin to wonder if your penchant for overcomplexity is because you are unable to make your putative argument without it.<<

It seems clear to me that 'S's belief that p is warranted' can be expressed as the conjunction (S believes p).(S's belief that p is warranted), which was your original idea, but you want to symbolise this now as a single proposition. I don't have a problem with that. It now falls to you to update your argument for circularity accordingly and I will respond to that in due course, although I see no benefit in this course given the logical equivalence. Or maybe we should postulate thin circular prisms on axles that we could attach to carriages as a means of transport...


>>I have replaced one logical proposition with a simpler one, nothing more nothing less.<<

Less clarity in the lack of reference to p, which I notice you fix below.

>>Well there is since you fail to consider the LHS of the biconditional; that is if you want to shoehorn 'S believes p' into the RHS of the biconditional then you have to also shoehorn it into the left, giving:

Bsp.KSp<=>BSP.BSp(W).p

and since BsP 'cancels out' nothing is served by including it. The burden of proof still lies with you!<<

You seem to be labouring under the misconception that a biconditional behaves like an equation and that introducing an entailed conjunction is like multiplying one side. That's just nonsense. Also, there is no 'shoehorning' going on, or have you forgotten the arguments /you/ offered detailing Russell's theory of Descriptions during the Onegin shot Lensky discussion? As for 'burden of proof', you have changed your mind and want to symbolise the argument differently. So it falls to you to update your argument doesn't it?

>>Because the two propositions may be equivalent and the outcome of the two truth tables may be equivalent but the two truth tables are not equivalent in complexity and I am unclear why you are so averse to demonstrating the validity and soundness of your putative argument using the simpler proposition/truth table rather than devoting a whole post to coming up with less than wholly persuasive arguments to avoid doing so.<<

What is your argument for circularity given your simplified formulation? Have I missed it? If you have not made it yet, then it is a bit rich for you to be calling for me to offer another counter argument isn't it? I have already demonstrated validity and soundness in a logically equivalent argument. It is no use trying to move the goal posts.


>>Methinks you protest too much.<<

Methinks you should get on and make your new argument for circularity. Unless, of course [cough] it is the same...
smiley - smiley

>>Simply resolved.

"S knows that p, therefore, if and only if S’s belief that p is warranted and p is true."

KSp<=>BSp(W).p<<

Splendid. And so much simpler smiley - ok

>>I was going to develop my argument here but, on reflection, better to do it separately.<<

I look forward to it.

ttfn


Methodology and ontology..Beyond the 'end of days'

Post 227

Bx4

hi psi

Apologies for the delay. I got involved in the subsidiary issue of whether in a biconditional as definition the case where the definiendum and the definiens a both true and the case where they are both false represent the same definition or not.

I am trying to do this from first principles and this is proving rather difficult because I have limited access to reference material here and Google is particularly useless when 'definition' is one of the search terms) so frequent trips to the university library have been necessary.

However nearly finished and will hopefully, after my usual rigorous proof reading smiley - winkeye, post in the next day or two.

bs


Methodology and ontology..Beyond the 'end of days'

Post 228

Psiomniac

Hi Bx4,

I look forward to the fruits of your research.

ttfn


Methodology and ontology..Beyond the 'end of days'

Post 229

Bx4

hi psi

Apologies. Unable to proofread until nnow. I loaded some 'critical updates' from Bloatware Inc. which completely buggered my system and it took me until now to diagnose and fix the problem.

Will proof read and hopefully post tomorrow.

bs


Methodology and ontology..Beyond the 'end of days'

Post 230

Bx4

Hi psi


This post is essentially a sidebar looking at your assertion that in a biconditioal definition, the case where both definiens and definiendum are both true and the case where both definiens and definiendum are both false, then both the true biconditionals that result are definitions of the same thing.


Operators



= equals

. conjunction

<=> material equivalence (iff)

<=>df definitional functor* (df a substitute for non-printing subscript)


A Universal quantifier (substitute for non-printing character)







Other conventions adopted;


{T} Truth value of 'true' in an element of a truth table(e.g. x{T} = 'The proposition x is true'


{F} 'Truth value of 'false' in an element of a truth table (e.g. x{F} = 'The proposition x is false.




if we take three propositions q, r and t linked by the biconditional operator;


(a) q<=>r.t*

then the truth table is for the biconditional is as follows:

(Row 1) q{T} (r.t){T} (q<=>r.t){T}

(Row 2) q{T} (r.t){F} (q<=>r.t){F}

(Row 3) q{F} (r.t){T} (q<=>r.t){F}

(Row 4) q{F} (r.t){F} (q<=>r.t){T}

Since Rows 2 and 3 both yield the truth value F for the biconditional they can be eliminated from the context in which the biconditional is a definition .


Moreover the truth table for the right hand side of the biconditional,the conjunction r.t, is:


(Row 2.1) r{T} t{T} (r.t){T}
(Row 2.2) r{T} t{F} (r.t){F}
(Row 2.3) r{F} t{T} (r.t){F}
(Row 2.4) r{F} t{F} (r.t){F}

But given that Merricks' biconditional definition states:


(A) 'S knows that p, therefore, if and only if S's belief that p is warranted and //P is true//'[emphasis added]

KSp<=BSpW.p{T}


then applying the same constraint we can eliminate the rows (2.2 & 2.4) in which the second proposition, t,in the conjunction r.t, is false from further consideration.

Therefore the original biconditional,q<=>r.t can be reformulated, using (Row 1) of the first truth table and (Row 2.1) from the second, as:

(b) [(q{T})<=>(r{T}).(t{T})]{T}

or, using (Row 4) of the first truth table and (Row 2.3) from the second, as:

(c) [(q{F})<=>(r{F}).(t{T})]{T}

Further more since:

'[D]efinitions have the form of [material] equivalences or identities; the symbol for [material] equivalence or identity sometimes accompanied by the abbreviation 'df' is said to be a definitional functor. The expression being defined, say A, on the left of the definitional functor is called the definiendum, and the expression use for defining, occurs on the right side [of the definitional functor] is called the definiens.'
{Logic from a Rhetorical Point of View,p.189; Witold Marciszewski; Walter de Gruyter & Co; 1994}

then the two biconditional (b) and (c) can, to show that they are definitions, be rewritten as:

(b1) [(q{T})<=>df(r{T}).(t{T})]{T}

and


(c1)[(q{F})<=>df(r{F}).(t{T})]{T}



So if q is treated as the definiendum and and r.t as the definiens, then in the definition q<=>r.t the truth value of the definiendum is dependent of the truth value, either true or false, of the definiens.

Therefore, in (a1) and (b1) above, given that t is invariably true, the truth value of the definiedum q is //wholly// dependent on the truth value, true or false, of r.

Clearly all of the above is dealing with the syntax of the biconditional as definition but it provides the basis for considering the semantics of a biconditional as definition.

Consider the, philosophically neutral, definition of an equilateral triangle:


(d) A closed plane figure is an equilateral triangle <=>df(it has three equal sides and three vertices)

Following Merricks definition if the second proposition of the definiens is //invariably// true, then (d) can be expressed in terms of (b1) and (c1) as:


(d.1) [(a closed plane figure is an equilateral triangle{T})<=>df(it has three equal sides{T}).(it has three vertices{T})]{T}

and


(d.2) [(a closed plane figure is an equilateral triangle{F})<=>df(it has three equal sides{F}).(it has three vertices{T})]{T}



Moreover, asserting the falsity of a proposition is logically equivalent to asserting the truth of its negation:


'Negation: The negation of a proposition is the denial of the truth of that proposition, which, in a truth-value logic (like the propositional calculus) is equivalent to the assertion of the proposition's falsity. If a proposition is true its negation is false, if it is false, its negation is true.'
(http://www.loyno.edu/~folse/symbtips.html)


Hence, the proposition 'a closed plane figure is an equilateral triangle is false' is logically equivalent to the negation 'a closed plane figure is a non-quadrilateral triangle is true'. Similarly the first proposition of the definiens, 'it has three equal sides is false' is logically equivalent to 'it has three non-equal sides is true'.

So (d.2) can be rewritten as:

(d.3)[(a closed plane figure is a non-equilateral triangle{T})<=>df(it has three non-equal sides{T}).(it has three vertices{T})]{T}


While the translation of (d.2) to (d.3) is not strictly necessary is serves to highlight the fact that while both the biconditional (d.1) and the biconditional (d.2)/(d.3) are both true definitions they are not definitions of the same thing.

The biconditional (d.1) is a definition of the set of all triangles that //are// equilateral triangles whereas (d.2)(d.3) is a definition of the set of of all triangles that //are not// equilateral triangles, that is, the set of all triangles that are either scalene or isosceles.


Applying this argument to Merricks', considerably less neutral, biconditional definition:



(f) S knows that p<=>df(S's belief that p is warranted).(p is true)

is true when:

(f.1) [(S knows that p){T})<=>df(S's belief that p is warranted{T}).(p{T})]{T}

and when

(f.2) [(S knows that p){F})<=>df(S's belief that p is warranted{F}).(p{T})]{T}


However, for clarity, given that to state a proposition is false is logically equivalent to stating the truth of its negation, then (f.2) can be written as:

(f.3) [(S does not know that p){T})<=>df(S's belief that p is not warranted{T}).(p{T})]{T}

Since p is invariably true the truth or falsity of the definiendum is wholly dependent on the truth or falsity of the first proposition of the definiens.

Again while both biconditional definitions are true they are not definitions of the same thing (f.1) is a definition of knowledge given that both propositions of the definiens are both true while (f.2)/(f.3) is a definition of ignorance (in the narrow epistemic sense).

So the definition (f.2)/(f.3) is irrelevant in the context of a definition in which S knows p and thus can be excluded from any discussion Merricks' definition. Only (f.1) needs to be considered.**




Notes:



* In a definition the universal quantifier is usually taken as implicit so

(Ax)(q<=>r.t) = q<=>r.t



**This argument holds, even if we accept, arguendo, Merricks somewhat problematic claim that (f.1) 'is a purely formal characterisation of warrant' since f2/f3 means that 'S belief that p is not warranted' which contradicts the claim, which seems fundamental to Merricks' 'characterisation'.



Methodology and ontology..Beyond the 'end of days'

Post 231

Psiomniac

Hi Bx4,

Sorry for the delay, I have been a bit busy.

>>if we take three propositions q, r and t linked by the biconditional operator;


(a) q<=>r.t*

then the truth table is for the biconditional is as follows:

(Row 1) q{T} (r.t){T} (q<=>r.t){T}

(Row 2) q{T} (r.t){F} (q<=>r.t){F}

(Row 3) q{F} (r.t){T} (q<=>r.t){F}

(Row 4) q{F} (r.t){F} (q<=>r.t){T}

Since Rows 2 and 3 both yield the truth value F for the biconditional they can be eliminated from the context in which the biconditional is a definition .


Moreover the truth table for the right hand side of the biconditional,the conjunction r.t, is:


(Row 2.1) r{T} t{T} (r.t){T}
(Row 2.2) r{T} t{F} (r.t){F}
(Row 2.3) r{F} t{T} (r.t){F}
(Row 2.4) r{F} t{F} (r.t){F}<<

So far, so good. But then you say:

>>(A) 'S knows that p, therefore, if and only if S's belief that p is warranted and //P is true//'[emphasis added]

KSp<=BSpW.p{T}


then applying the same constraint we can eliminate the rows (2.2 & 2.4) in which the second proposition, t,in the conjunction r.t, is false from further consideration. <<

and here I disagree. It seems clear to me that Merricks has added the 'is true' for readability, but Ramsay's Ladder applies and thus the correct formulation in the manner of the above proposition is:

KSp<=BSpW.p

So I'm afraid the next sections of your post, excluding row 4 of the truth table as they do, rest on the error of assigning T to the propositional variable p, making it invariant for the definition. p would thus be a constant, which makes no sense.

>>Therefore, in (a1) and (b1) above, given that t is invariably true, the truth value of the definiedum q is //wholly// dependent on the truth value, true or false, of r. <<

The nonsensical nature of this interpretation, where a variable can be assigned an /invariant/ truth value, rendering it a constant, can be seen when you consider the semantics below:

>>Clearly all of the above is dealing with the syntax of the biconditional as definition but it provides the basis for considering the semantics of a biconditional as definition.

Consider the, philosophically neutral, definition of an equilateral triangle:


(d) A closed plane figure is an equilateral triangle <=>df(it has three equal sides and three vertices)

Following Merricks definition if the second proposition of the definiens is //invariably// true, then (d) can be expressed in terms of (b1) and (c1) as:


(d.1) [(a closed plane figure is an equilateral triangle{T})<=>df(it has three equal sides{T}).(it has three vertices{T})]{T}

and


(d.2) [(a closed plane figure is an equilateral triangle{F})<=>df(it has three equal sides{F}).(it has three vertices{T})]{T}<<

Why would anyone, (or Merricks in the case of warrant), assume that the three vertices condition is /invariably true/? Perhaps what has happened here to make it seem even vaguely plausible is that the three vertices property is entailed by the three sides one. However, the analagous case for warrant does not hold a priori.

To help illustrate the semantics further, and clarify the point above, let's consider another definition:

A closed plane figure is a square iff it has four equal sides and all its internal angles are right angles

Then symbolizing this as s <-> e.r, where s = is square, e = four equal sides and r = all internal angles are right angles, we can see that it makes no sense to hold as invariantly true the condition that the internal angles are right angles. That would be to ignore the existance of the rhombus!

This is still a true definition of a square in cases where either e or r or both are F so long as s is also F. In the case of a rhombus, s = F, e = T, r = F. The fourth row of the truth table applies, the definition has done its job by saying 'it is false that this is a square.'


>>Moreover, asserting the falsity of a proposition is logically equivalent to asserting the truth of its negation: <<

I agree. But the rest of your post, including:

>>Since p is invariably true the truth or falsity of the definiendum is wholly dependent on the truth or falsity of the first proposition of the definiens.

Again while both biconditional definitions are true they are not definitions of the same thing (f.1) is a definition of knowledge given that both propositions of the definiens are both true while (f.2)/(f.3) is a definition of ignorance (in the narrow epistemic sense).

So the definition (f.2)/(f.3) is irrelevant in the context of a definition in which S knows p and thus can be excluded from any discussion Merricks' definition. Only (f.1) needs to be considered.** <<

..rests on the error mentioned above, of treating 'p is true' as an invariant assigning of p as true, which as I have illustrated above, is not sensible.

The last part also seems to follow from a confusion about the negation of propositions in contrast to assigning the value F to one or more of its variables.

Returning the the definition of a square, we considered the case where the internal angles were not 90 degrees, and even though it had four equal sides, it is not a square. The definition of square is true in this case, since it is false that this is a square. However, the definition used was:

s <-> e.r

which is NOT the same as using the logically equivalent definition of a non-square:

¬s <-> ¬(e.r)

So whilst I agree that semantically there is a difference between the logically equivalent definition of an X and the definition of a non-X, this issue does not arise when considering circularity in Merricks' argument.

I hope that makes sense,

ttfn.


Methodology and ontology..Beyond the 'end of days'

Post 232

Bx4

hi psi

>>Sorry for the delay, I have been a bit busy.<<

I'm in a similar position. Wall to wall visitors who need to be entertained. They are off tomorrow so I will essay a reply over the weekend

bs


Methodology and ontology..Beyond the 'end of days'

Post 233

Psiomniac

Hi Bx4,

I look forward to your reply.

ttfn


Methodology and ontology..Beyond the 'end of days'

Post 234

Bx4

hi psi

Slight delay due to a impulsive decision to travel to Edinburgh for Festival

bs


Methodology and ontology..Beyond the 'end of days'

Post 235

Psiomniac

Hi Bx4,

Ah, I did wonder whether you might have. Seen anything good?

ttfn


Methodology and ontology..Beyond the 'end of days'

Post 236

Bx4

Hi psi

Fringe is proving more OTT than ever with way too many shows. Strangely very little Jazz perhaps because there is now a Edinburgh International Jazz Festival earlier in the year. I did catch a group called the Robert Glasper trio who were OK. Tomorrow we are going to see Antonio Forcione and Adriano Adewale. I'v seen the former at the Festival before (I recall your saying you had gigged with him) though I don't know Adewale. There is also a one-off show with the Sun Ra (deceased) Arkestra which I might go to though I don't know their work.

Managed to catch a few Blues gigs, a Scottish band called Main Street Blues who were pretty good and a blues pianist called Daniel Smith. On the 'world' music front a surfeit of Japanese drum groups, Sumurai Drum Ikki, tonight and going to see Japan Marvellous Drummers tomorrow night. I've liked japanese drumming since I saw Ondekoza a few years back. One or two interesting programes of Chinese music at the Confucius Institute next week.

Always a bit of a sucker for Bach when played on the harpsichord so I liked Richard Eggar's recital and also Tamar Fejes recital of his solo violin works. Went on spec to what I thought was a recital of Stockhausen's Tierkreis but it proved not be but rather an 're-interpration' by someone called Haworth Hogdkison celebrating

Tierkreis fortieth anniversary played on organ and percussion. Also feature works by Leguay, Weir, Kurtag and Messiaen. Slightly odd but quite enjoyable.

Saw two quite spectacular events one called 'the Harmonium Project' an open air sound and light show which featured a recital of John Adams 'Harmonium'and a computer generated light show based on the performance and projected onto the Usher Hall.

http://www.ed.ac.uk/polopoly_fs/1.170274!/fileManager/Harmonium%2059%20Productions%20%20400x200.jpg


Another, albeit slightly less successful, was 'AniMotion' a collaboration between Evelyn Glennie and the artist Maria Rud which was projected onto Heriot Watt High School though I believe the projection used avideo made a few years ago.


Mostly avoided comedy a revue shows except for Mark Steele who was brilliant. Some plays among which Elspeth Turner's 'Spectre Town', rooted in the North East Scotland 'Bothy Ballad' tradition and Robert Le Page's '887'. Also some interesting exhibitions notably of works by M C Escher and Roy Lichtenstein.


I am beginning to get quite fagged out by the 'festival experience. demonstrating that I am no longer who I

once was so a less intensive last week is indicated.smiley - winkeye


Anyhow returning to things filosofickal: I'm not really clear as the statements you make over ' p is true' in

Merricks 'purely formal characterisation' and until I can get you to clarify them I can't really respond to

your later comments

First you say:

'It seems clear to me that Merricks has added the 'is true' for readability...'

Perhaps you could clarify what you mean by 'readability' and why you think Merricks added 'is true'

particulaly to p but not to KSp and BSpW?

However, to state the obvious, in propositional logic any arbitrary proposition p can, exclusively, take one

one of two truth values. 'true' or 'false' and in asserting 'p is true' Merricks is excluding 'p is false'

from consideration which does more than improve mere 'reaability'.

You continue:

'...but Ramsay's Ladder applies and thus the correct formulation in the manner of the above proposition is:

KSp<=BSpW.p'

This is something of a red herring. 'Ramsey's Ladder' is not a principle (or rule) of propositional logic but

rather rather an argument made in support of a (deflationary) of truth.

Again, to restate the obvious,in propositional logic any arbitrary proposition p can, exclusively, take one

one of two truth values. 'true' or 'false'.

So for p we have either (XOR):

(a) p is true

and

(b) p is false

So contrary to Blackburn, to'p is true' or 'p is false' is not the same as saying p.


You then say:

'So I'm afraid the next sections of your post, excluding row 4 of the truth table as they do, rest on the error of assigning T to the propositional variable p, making it invariant for the definition.'

I am merely following Merricks here in making the the //truth value// of p invariant not as you suggest the //propositional variable// p itself invariant.

You then claim:

'p would thus be a constant, which makes no sense.'

The proposition p is a variable, in that it could be any arbitrary proposition, but, because Merricks has assigned the truth value 'true' to it, its truth condition is a constant. Please clarify why you think this 'makes no sense'.

You then say:

'The nonsensical nature of this interpretation, where a variable can be assigned an /invariant/ truth value, rendering it a constant.

I think you are making a category error here; a propositional variable p becomes a 'constant' when we assign a meaning to it, as in:

(1) p = the plane figure is an equilateral triangle

Assigning a (truth) value to a prpositional does not make it a constant as assigning a (numerical) value an algebraic variable.

(2)x = 5

So when Merricks says 'p is true' it is not the propositional variable p that becomes a constant but rather its truth value that does. I am unclear why you think this is 'nonsensical'.

bs


Methodology and ontology..Beyond the 'end of days'

Post 237

Psiomniac

Hi Bx4,

Sounds like you've been catching some interesting stuff, I hope that continues this week.

So, on to the philosophical:

>>'It seems clear to me that Merricks has added the 'is true' for readability...'

Perhaps you could clarify what you mean by 'readability' and why you think Merricks added 'is true'

particulaly to p but not to KSp and BSpW? <<

Yes, I'll try to make this clear, it might help also to compare with the definition of a square example, since this demonstrates that assigning T as invariant to one of the variables is counter to the semantics.

First in terms of readability, compare 'S knows that p, therefore, if and only if S's belief that p is warranted and p is true', with 'S knows that p, therefore, if and only if S's belief that p is warranted and p'

i hope that you can see that the latter, whilst making sense to logicians, sacrifices readability. An alternative might have been '...S's belief that p is warranted and it is the case that p', but 'p is true' is less cumbersome. The reason that 'is true' has not been added to the other clauses is clear if you try it:

'S knows that p is true, therefore, if and only if S's belief that p is warranted is true and p is true'

..which is confusing and ambiguous because of the structure of English. For example, in the second clause, what we mean is that it is true that S's belief that p (is true) is warranted. So I hope you can see that Ramsey's Ladder, rather than being a red herring, explains why 'is true' is omitted, except in the case when readability is compromised.

Compare now with the square example, we could replace 'A closed plane figure is a square iff it has four equal sides and all its internal angles are right angles' with 'A closed plane figure is a square [is true] iff it has four equal sides [is true] and all its internal angles are right angles [is true]'

In this case, Ramsey's Ladder allows us to remove all the 'is true' modifiers to aid readability. This is because all the propositional variables have semantic content, whereas in Merricks' definition, the only one that doesn't have any is 'p'. So we can assert S knows p, p is warranted etc, but just to assert p requires 'is true' for readability.

Also, let's remember that in the square example it makes no sense to ascribe an /invariant/ truth value to one of the variables for the purposes of the definition. this is the case in Merricks' definition also, but perhaps you could clarify why you think it does?

The above brings us to the question of assigning a value to a variable. You make a good point that assigning a value to a variable does not make it a constant, I concede that. For example if we are considering the general straight line y = mx + c, we might consider the case when x=2, but x is still a variable, whereas c and m are constants. However, Merricks sets up the definition with p as a /variable/ from the beginning. So setting p = T would be like setting up a scenario y = mx + c but only considering x = 2. So why set up a line scenario and then only consider a point? How is the line relevant? Perhaps you could clarify why you think it makes sense for Merricks to set up a variable and then only consider one value?

>>So contrary to Blackburn, to'p is true' or 'p is false' is not the same as saying p. <<

Blackburn does not say that 'p is false' is the same as saying p. I think this is an old error of yours. You seem to confuse the cases of asserting p and yet p being false, and asserting ¬p. These are distinct, despite the logical equivalence of asserting 'p is false' and asserting '¬p'.

Must dash, but I hope that clarifies sufficiently for you to respond to my later points in the previous posts.

ttfn


Methodology and ontology..Beyond the 'end of days'

Post 238

Bx4

Hi psi

I started to suffer from 'Fringe Fatigue' last weekend so I only took in the Forcione gig (fine) and a couple of events at the Confucius Institute oterwise pottering about with friends.


>>Must dash, but I hope that clarifies sufficiently for you to respond to my later points in the previous posts.<<

Not really. Before responding to your post in detail there is a point that I want to clarify in relation to 'readability' and 'Ramsey's Ladder'

Accepting, arguendo, Ramsey's redundancy theory of truth then as I understand your argument it is as follows:

(a) If a proposition has no explicit 'semantic content' then 'is true' is retained but only to improve 'readability'.

(b) If a proposition has some explicit 'semantic content' the 'is true' does not improve 'readability' and is omitted because again, stricly, it is redundant.

However, since 'is true' is retained or omitted as a matter of 'readability', then //strictly// in terms of Ramsey's redundancy theory 'is true' is redundant in both contexts.. Therefore:

(i)To assert 'p is true' is simply to assert 'p' because, strictly, 'is true' is redundant.

(ii)To assert 'S knows that p is true' is simply to assert 'S knows that p' because, strictly, 'is true' is redundant.


(iii)To assert 'S's belief that P is warranted is true' is the same as to assert 'S's belief that p is warranted' because strictly, 'is true' is redundant.

Is this is a fair summary of your argument about 'readability'?

If so, then presumably it follows that if the retention or omission of 'is true' is simply a matter of 'readability' then the truth values of the propositions (i), (ii) and (iii) are unaffected whether 'is true' is retained or omitted?

bS


Methodology and ontology..Beyond the 'end of days'

Post 239

Psiomniac

Hi Bx4,

Your summary is fair as far as it goes, but I am not sure you have addressed my point about how word order in English further complicates things. So in order to add the redundant 'is true' and preserve the meaning, we would have to use some different constructions. So whereas:

i) 'S knows that p, therefore, if and only if S's belief that p is warranted and p is true'

is more readable than the logically equivalent:

ii) 'S knows that p, therefore, if and only if S's belief that p is warranted and p'

If we want to add redundant affirmations of the truth of each proposition we would have to say something like:

iii) 'It is true that S knows that p, therefore, if and only if it is true that S's belief that p is warranted and p is true'

which avoids the ambiguity I mentioned.

So if you make those modifications then I think it will be a fair summary.

ttfn


Methodology and ontology..Beyond the 'end of days'

Post 240

Bx4

hi psi

>>Your summary is fair as far as it goes, but I am not sure you have addressed my point about how word order in English further complicates things.<<

That is because I don't see the complication you claim. In 'Facts and Propositions'(p.4) Ramsey makes that there is a distinction between the restricted English of his redundancy theory and that of 'ordinary' English when he says:

'... the propositional function p is true is simply the same as p, as e.g. its value 'Caesar was murdered is true' is the same as 'Caesar was murdered...We have in English to add 'is true' to give the sentence a verb, forgetting that ' p ' already contains a (variable) verb.'.

So it is by no means clear to me why a professional philosopher like Merricks would feel the need to add 'is true' merely to give p an unnecessary verb. So it seems plausible that he might have a reason other than merely stylistic ones.

>So in order to add the redundant 'is true' and preserve the meaning, we would have to use some different constructions. So whereas:

i) 'S knows that p, therefore, if and only if S's belief that p is warranted and p is true'

is more readable than the logically equivalent:

ii) 'S knows that p, therefore, if and only if S's belief that p is warranted and p'

If we want to add redundant affirmations of the truth of each proposition we would have to say something like:

iii) 'It is true that S knows that p, therefore, if and only if it is true that S's belief that p is warranted and p is true'

which avoids the ambiguity I mentioned.


So if you make those modifications then I think it will be a fair summary<<

Well not quite since you have dropped 'is true' from the occurrence of p in the first two propositions so for absolute clarity surely iii) should be :

iv) It is true that S knows that p is true, therefore, if and only if it is true that S's belief that p is true is warranted and p is true'.

The 'redundant affirmations' in iii) and iv) seem to serve the purpose removing any ambiguity as to the truth values of //all// three propositions.

bs


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