h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Hi Bx4,

Sorry for the delayed reply, I am rather time poor at the moment but this should change soon.

It looks as though we might be closer to agreement given your last. In that regard though it seems to me that our original objection to my setting of truth values no longer applies, in which case I can't see how your argument for circularity survives.

Maybe you can say more on this, but I'll reply more fully after Wednesday, I should have more time then.

ttfn

Sorry, 'our original objection' should read 'your original objection'.

hi psi

>>Sorry for the delayed reply, I am rather time poor at the moment but this should change soon.<<

No problem. I'm still pretty time poor myself at the moment - The act of retiring being more prolonged that I had anticipated.

>>It looks as though we might be closer to agreement given your last. In that regard though it seems to me that [y]our original objection to my setting of truth values no longer applies...<<

Sorry my objection still stands. I have no problem in agreeing that truth table can be applied to an arbitrary biconditional, say p<=>r.s, but when this biconditional is characterised as a definition I don't think it can. I have already given two links (see earlier posts) which make it clear that if the biconditional is a true definition then necessarily both the definendum,(p), and the definiens,(r.s) are true.

Therefore, the use of truth tables which rely on varying the truth values of (p) and (r.s) are irrelevant to my argument.

My intention in discussing the context where (p) and (r.s) were both false is that this equivalent to ¬p<=>¬(r.s) which may well be a definition but clearly a different one from p<=>r.s.

>>...in which case I can't see how your argument for circularity survives.<<

If you accept my premise that, necessarily, if a biconditional represents a true definition then necessarily both the definiendum and its definiens are true then I can show how my argument for circularity survives.

bs

Hi Bx4,

Well I do hope that the prolonged retirement process isn't of the irksome kind.

It seems that we have run up against the same nub of disagreement here. Your links say that in a definition both the conditional and its converse must be true. This is different from your argument that in:

1) p <-> r.s

both the LHS and the RHS must be true. Your argument is mistaken because the conditional and converse can be true when p, r and s are false. This does not constitute a different biconditional.

It seems to me that you have confused 'the definendum,(p)' with the conditional, and the 'definiens,(r.s)' with the converse. They are not the same and it is the latter in each pair that is necessarily true in a definition.

If you could address this issue directly it might help.

ttfn

hi psi

I am still unclear as to whether you accept that (see the two links I gave earlier) in the specific context of a biconditional, say p<->q.r, functioning as a //definition// not only is the biconditional true but the proposition on the left hand of the statement, p, and the conjoined proposition,q.r. on the right hand side of the biconditional.**


' Your argument is mistaken because the conditional and converse can be true when p, r and s are false. This does not constitute a different biconditional.'

It may not constitute a different biconditional but I consider that it constitutes a different //definition// as consideration of the logical semantics shows.

Consider a concrete example that contains no problematic philosophical terms:

(a) A figure is a square if and only if all four sides are congruent and all four angles are congruent. ***

Clearly for this definition to be true both its LHS and its RHS are true. However if we consider the case where the LHS and RHS are false we have:

(a.1) It is false that the figure is a square if and only if it is false that all four sides are congruent and all four angles are congruent.

or

(a.2) The figure is not a square if and only if it is false that all four sides are congruent and all four angles are congruent.

Then you have to demonstrate that (a.1) is also an equivalent true definition of a square rather than simply a true but non-definitional biconditional.

' It seems to me that you have confused 'the definendum,(p)' with the conditional, and the 'definiens,(r.s)' with the converse. They are not the same and it is the latter in each pair that is necessarily true in a definition.

I assume you accept in that in the definition**** (a), the definiendum **** is 'the figure is a square and the definiens**** is 'all four sides are congruent and all four angles are congruent'

I think the mistake is yours in that you are confusing the infelicitous naming of the of the definiendum as the conditional and the definiens as the converse as implying the have a logical operator embedded in them. As I have already pointed out the not uncommon usage is to replace conditional and converse in a biconditional with the more neutral alternatives 'left hand side' and 'right hand side'; e.g.

https://www.google.co.uk/search?q=biconditional+left+hand+side&ie=utf-8&oe=utf-8&gws_rd=cr&ei=KuwOVerWMs_taJWagLAB

Moreover surely if (a) is a definition of a square then if the definiens is as you agree necessarily true the surely the definiendum is also necessarily true else the definition would be false.

'If you could address this issue directly it might help.

I think I have the burden of proof now lies with you.

bs



**I prefer to use the convention 'left hand side' and 'right hand side' (e.g., see An Introduction to Metalogic: Aladdin M. Yaqub, Broadview Press, 2014, pps. 18,19) because while the biconditional expansion:

(i) (p->q.r).(q.r->p)

results in a conditional and conjoined converse conditional it is misleading to name the propositions p and q.r in the unexpanded biconditional as if they contained embedded operators.

***See the account of polygons, quadrilaterals, parallelograms and the square here:

http://www.andrews.edu/~calkins/math/webtexts/geom02.htm

http://www.andrews.edu/~calkins/math/webtexts/geom06.htm


**** definition: A statement of the exact meaning of a word, especially in a dictionary.

definiendum: A word, phrase, or symbol which is the subject of a definition...which is introduced into a logical system
by being defined.

definiens: A word, phrase, or symbolic expression used to a word or symbol into a logical system by providing a
statement of its meaning.

Hi Bx4,

In your #143 you offered this:

>>'A biconditional statement can be either true or false. To be true, //BOTH// the conditional statement and its converse must be true. This means that a true biconditional statement is true both "forward" and "backward". //All definitions// can be written as //true// biconditionals' [emphases added] <<

So I had thought we'd agreed that a biconditional can serve as a definition so long as both the conditional and its converse are true.

But for a while now you seem to have confused this with the following:

>>(a) A figure is a square if and only if all four sides are congruent and all four angles are congruent. ***

Clearly for this definition to be true both its LHS and its RHS are true.<<

The above is just wrong. According to your #143 (with which I agree), both the conditional and its converse must be true. This is /not/ the same as and nor does it entail the LHS and RHS being true.

I have explained this in detail in previous posts but I am happy to go over the logic again if you like.

So could you please try to directly address the issue that the conditional and its converse /do not/ map directly onto the LHS and RHS? Until you have done this, the burden is still with you I'm afraid.

>>I am still unclear as to whether you accept that (see the two links I gave earlier) in the specific context of a biconditional, say p<->q.r, functioning as a //definition// not only is the biconditional true but the proposition on the left hand of the statement, p, and the conjoined proposition,q.r. on the right hand side of the biconditional.**<<

See above. I do not accept this and nor did your links say this. Rather, I agree with your #143 statement, which says that for a definition the conditional and its converse must be true. In the above, the conditional is p->q.r and the converse is q.r->p

I hope that clarifies,

ttfn

Hi psi
Been off on a sailing holiday - just got back so haven't had a chance to read your latest yet will reply as soon as I have.

bs

Hi Bx4,

Sounds fun, I look forward to your response when you get chance.

ttfn

hi psi

Sailing was great.



>>So I had thought we'd agreed that a biconditional can serve as a definition so long as both the conditional and its converse are true.<<

Before getting further in to your argument. I want to clarify what we agree on.

Given the intensional biconditional definition D:


(D) For any x, is a square if and only if x has all four sides congruent and x has all four angles congruent.


then if D is a true biconditional definition then its component propositions

(p) x is a square

(q) x has all four sides congruent

(r) x has all four angles congruent

are all necessarily true.

So are we are in agreement this far and if not why not?

bs

Hi Bx4,

This is the nub of our disagreement, yes:

>>Before getting further in to your argument. I want to clarify what we agree on.

Given the intensional biconditional definition D:


(D) For any x, is a square if and only if x has all four sides congruent and x has all four angles congruent.


then if D is a true biconditional definition then its component propositions

(p) x is a square

(q) x has all four sides congruent

(r) x has all four angles congruent

are all necessarily true.

So are we are in agreement this far and if not why not? <<

No, we are not in agreement. The reason, as I said in my previous post, is that I agree with what you originally said, which is that for a true definition expressed as a biconditional, the /conditional/ and its /converse/ must be true.

So before we proceed, perhaps you could clarify why you have changed your view? Is it that you think the above is equivalent in some way? If so, how?

ttfn

hi psi

>>No, we are not in agreement. The reason, as I said in my previous post, is that I agree with what you originally said, which is that for a true definition expressed as a biconditional, the /conditional/ and its /converse/ must be true.<<

I have not changed my position, except in terms of terminology. The problem may lie in different interpretations of the terms conditional and converse in a biconditional as distinct from the more general meaning of the terms.

I thought I had clarified earlier when I said that the older terminology might be read as implying that some property of conditionality inheres in the propositions on both sides of the biconditional operator.

I think this is nonsensical which is why why I prefer the more neutral terminology being adopted by some modern authors, namely (the proposition on) the left hand side of expression and (the proposition on) the right hand side of the expression.***

I think the confusion arises because the biconditional expansion of P<->Q; P->Q.Q->P gives a conditional and a (converse) conditional. However, the property of conditionality lies in the operator, ->. and not in the propositions P and Q.****Similarly in the biconditional the property of biconditionality lies in the operator and not in the propositions P and Q.

>>So before we proceed, perhaps you could clarify why you have changed your view? Is it that you think the above is equivalent in some way? If so, how? <<

obviously the truth functionality of the biconditional is invariant with respect to either way of naming the proposition P and Q. For the reason given I prefer the alternative usage.


Perhaps now that I have clarified this you could address the substantive argument in my last post.


***I have already cited an example of this usage in Yaqub but it can also be found in number of other authors; e.g.:

'In the language of PL where 'P' is a proposition and 'Q' is a proposition, a proposition of the form:

P<->Q

is called a biconditional (the material biconditional). The symbol <-> is a truth functional operator.... The proposition to the left hand side of the double arrow is called the left hand side of the biconditional, while the proposition to the right hand side of the double arrow is called the right hand side of the biconditional. That is


Left Hand Side Right Hand Side

P <-> Q

The truth function corresponding to <-> is the following:

Biconditional = df. If the truth value input of the propositions are identical the complex proposition is true. If the truth value inputs differ the complex proposition is false.'

(Symbolic Logic, Sytax, semanics and Proof: David Agler; Rowman Littlefield; P. 51)

****Note as I said earlier I now take the view that the biconditional expansion is not necessary for my argument; the simple biconditional will do.

bs

Hi Bx4,

>>I have not changed my position, except in terms of terminology. The problem may lie in different interpretations of the terms conditional and converse in a biconditional as distinct from the more general meaning of the terms.<<

In the biconditional p<->q the conditional is p->q and the converse is q->p. I made this case a few posts ago and don't know of any different interpretations apart from, perhaps, yours.

>>I thought I had clarified earlier when I said that the older terminology might be read as implying that some property of conditionality inheres in the propositions on both sides of the biconditional operator.<<

I have consistently advocated that the conditional is p->q and the converse is q->p.

>>I think this is nonsensical which is why why I prefer the more neutral terminology being adopted by some modern authors, namely (the proposition on) the left hand side of expression and (the proposition on) the right hand side of the expression.***<<

The proposition on the LHS of p<->q is p. The proposition on the RHS is q. I agree on this terminology. However the conditional is p->q and the converse is q->p. So these are terminologies for different things.

>>I think the confusion arises because the biconditional expansion of P<->Q; P->Q.Q->P gives a conditional and a (converse) conditional. However, the property of conditionality lies in the operator, ->. and not in the propositions P and Q.****Similarly in the biconditional the property of biconditionality lies in the operator and not in the propositions P and Q.<<

See above.

>>obviously the truth functionality of the biconditional is invariant with respect to either way of naming the proposition P and Q. For the reason given I prefer the alternative usage.<<

I am fine with your system of naming P and Q. However, 'the conditional' does not name P or Q, it is a name for 'P->Q'.

>>Perhaps now that I have clarified this you could address the substantive argument in my last post.<<

It seems to me that your substantive point rests on a misconception arising from your original conflation of 'conditional' with one of the propositions. Perhaps this is because the terminology is similar to 'consequent', I'm not sure. However, both logically and semantically your example serves as a true definition even when both the LHS and RHS of the biconditional are false. To assume they must necessarily be true is incorrect. Hence my argument against circularity stands, since your objection to my setting of some truth values to false fails for the reasons described above.

If this doesn't clarify sufficiently, I can look at a semantic interpretation of your example?

ttfn

hi psi
Sorry for the delay. I have been a bit poorly of late.


Changing the order of your replies somewhat.

>>I have consistently advocated that the conditional is p->q and the converse is q->p.<<

Indeed but neither appear in the unexpanded by conditional.

>>In the biconditional p<->q the conditional is p->q and the converse is q->p.<<

Except that neither the conditional p->q nor the converse conditional q->P appear 'in' the unexpanded biconditional

>>>> I made this case a few posts ago and don't know of any different interpretations apart from, perhaps, yours.<<

My interpretation is that neither the conditional p->q nor the converse (conditional) q->p appear /'in'/ unexpanded biconditional p<=> as you state but only in the biconditional expansion p->q.q->p

>>It seems to me that your substantive point rests on a misconception arising from your original conflation of 'conditional' with one of the propositions<<

As opposed to your metaphysical claim that the conditional and converse are 'In the biconditional'


>> I'm not sure. However, both logically and semantically your example serves as a true definition even when both the LHS and RHS of the biconditional are false. To assume they must necessarily be true is incorrect. Hence my argument against circularity stands, since your objection to my setting of some truth values to false fails for the reasons described above.<<

Given the specific example:

'A figure is a square if and only if all four sides are congruent and all four angles are congruent.'

Perhaps you can give an account in terms of the semantics (meaning) rather syntax (logical structure) of the biconditional to explain how if both sides of the bicomditional are false it can be a true definition of a square.

Completely unrelated to this. I have been transferring my vinyl and CDs to my Brennan an I noticed for the first time, despite having played it quite a few times, the typo on the cover!!

bs

Hi Bx4,

Sorry to hear you haven't been well, I hope you are feeling better.

I'll reply in two parts, giving a semantic treatment of your example in the next reply.

First I want to deal with your rather odd notion that I am making some kind of metaphysical claim in dealing with the conditional and it's converse. First, it was you who introduced the assertion that a biconditional used as a true definition can function in this role only if the conditional and it's converse are both true. Were you making a metaphysical claim? I think not. Rather, you are using logical equivalence. You even spell this out in your #141:

>>P1 KcP<=>BcP.WP.P

which is equivalent to:

P1a (KcP->BcP.WP.P).(BcP.WP.P->KcP) <<

Perhaps you could clarify why you have suddenly construed an equivalence that you yourself introduced as 'metaphysical'?

ttfn

hi psi

>>Sorry to hear you haven't been well, I hope you are feeling better<<
Yes. Thanks for asking. Nothing too serious just the consequences of increasing decrepitude.

>>First I want to deal with your rather odd notion that I am making some kind of metaphysical claim in dealing with the conditional and it's converse.<<

I was not being wholly serious.

>>First, it was you who introduced the assertion that a biconditional used as a true definition can function in this role only if the conditional and it's converse are both true.<<
In the expanded biconditional

Were you making a metaphysical claim? I think not. Rather, you are using logical equivalence. You even spell this out in your #141:

>>P1 KcP<=>BcP.WP.P

which is equivalent to:

P1a (KcP->BcP.WP.P).(BcP.WP.P->KcP)<<<<

Indeed but to assert equivalence is not the same as to assert that the conditional and its converse are'in the biconditional'.

>>I'll reply in two parts, giving a semantic treatment of your example in the next reply<<

I look forward to it.

Just o clarify. We have no disagreement over the truth value of a /non-definitional/ biconditional being true either when both the LHS and the RHS are true or when both the LHS and RHS side are false.

Where we disagree is over whether the latter case has any relevance when we are dealing with a /definitional/ biconditional.

Using the biconditional definition of a square avoids any of the philosophical complication of Merrick's 'purely formal characterisation of warrant'. So I would suggest we begin with the former before moving on to the latter.

bs

Hi Bx4,

>>I was not being wholly serious. <<

Indeed.

>>Indeed but to assert equivalence is not the same as to assert that the conditional and its converse are'in the biconditional'. <<

I wasn't making that assertion, or perhaps you could say I wasn't being wholly pedantic By saying 'in the biconditional p <-> q.r' I simply mean to refer to that biconditional and the conditional and converse statements that, when conjoined, are logically equivalent to it. But 'in' was shorter.

>>Just o clarify. We have no disagreement over the truth value of a /non-definitional/ biconditional being true either when both the LHS and the RHS are true or when both the LHS and RHS side are false. <<

Yes, we agree on this.

>>Where we disagree is over whether the latter case has any relevance when we are dealing with a /definitional/ biconditional. <<

That's right. I agree with the quote (and link) in your #143, here it is:

>>'A biconditional statement can be either true or false. To be true, //BOTH// the conditional statement and its converse must be true. This means that a true biconditional statement is true both "forward" and "backward". //All definitions// can be written as //true// biconditionals' [emphases added] <<

There we have it, both the /conditional/ statement and its converse have to be true. The subsequent reference to being true both forward and backward makes this completely clear. In other words, for:

p <-> q.r

to work as a true definition, both the conditional statement:

p -> q.r (forward)

and p <- q.r (backward), equivalent to q.r -> p

must be true. Your link supports this interpretation.

>> https://docs.google.com/presentation/d/1JVSuRiX06kpJpdtakIkHjyy0Tj4uGdif-QHglEGv8Iw/edit?pli=1#slide=id.i57 <<

just a reminder of your invocation of the expanded equivalent form:

>>It therefore follows that Merricks:

(a)KcP<->BcP.WP.P

since it is a definition of knowledge and hence from the above must be true. I think therefore I am justified in expressing (a) as:

(b)(KcP->BcP.WP.P).(BcP.WP.P->KcP) <<

I think this disagreement happened because you confused the conditional statement and its converse with the statements on the LHS and RHS of the biconditional. The conditional statement p -> q.r /can/ be true when both p and q.r are false. similarly for the converse. Hence the conditions for a true definition are met when the LHS and RHS of the unexpanded biconditional are both false. In other words, the 'false on both sides' condition /is/ relevant.

>>'A figure is a square if and only if all four sides are congruent and all four angles are congruent.'

Perhaps you can give an account in terms of the semantics (meaning) rather syntax (logical structure) of the biconditional to explain how if both sides of the bicomditional are false it can be a true definition of a square.<<

So now we turn to the semantics. A definition of a square must include all squares and it must /exclude/ all non-squares. That's what the 'iff' does.

In particular, the property of the biconditional being true when both LHS and RHS are false can be understood to mean that it is the case that if you encounter a shape and it is false that it is a square, then it is false that all four sides are congruent and all four angles are congruent. If it is false that all four sides are congruent and all four angles are congruent, then it is false that it is a square. But the /definition/ is true.

I hope that clears up the issue.

ttfn

hi psi

>>I wasn't making that assertion, or perhaps you could say I wasn't being wholly pedantic<<

What some call pedantry others call precision.

>> But 'in' was shorter.<<
But also inaccurate.

>>Yes, we agree on this.<<

Good

>>Yes, we agree on this.

>>Where we disagree is over whether the latter case has any relevance when we are dealing with a /definitional/ biconditional. <<

That's right. I agree with the quote (and link) in your #143, here it is:

>>'A biconditional statement can be either true or false. To be true, //BOTH// the conditional statement and its converse must be true. This means that a true biconditional statement is true both "forward" and "backward". //All definitions// can be written as //true// biconditionals' [emphases added] <<

There we have it, both the /conditional/ statement and its converse have to be true. The subsequent reference to being true both forward and backward makes this completely clear. In other words, for:

p <-> q.r

to work as a true definition, both the conditional statement:

p -> q.r (forward)

and p <- q.r (backward), equivalent to q.r -> p

must be true.<<

I disagree with your interpretation of 'forward' and backward' as (necessarily?) requiring recourse the the biconditional expansion since I don't think:

'A biconditional statement can be either true or false. To be true, //BOTH// the conditional statement and its converse must be true. This means that a true biconditional statement is true both "forward" and "backward". //All definitions// can be written as //true// biconditionals' [emphases added]

says this. A simpler interpretation is that 'forward' means

p<=>q.r

and 'backward' means

q.r<=>p

>>I think this disagreement happened because you confused the conditional statement and its converse with the statements on the LHS and RHS of the biconditional. The conditional statement p -> q.r /can/ be true when both p and q.r are false. similarly for the converse. Hence the conditions for a true definition are met when the LHS and RHS of the unexpanded biconditional are both false. In other words, the 'false on both sides' condition /is/ relevant<<

Indeed though I later changed my position when I said that I ha realised that I disd not need the biconditional expansion to support my argument and was happy to use the unexpanded biconditional where the terms 'conditional' (LHS) and converse (RHS) do not have the same meaning as your usage above.

In the biconditonal 'conditional' simply means the proposition or propositions to the left hand of the biconditioal operator and 'converse' simply means the proposition or propositions to the right the biconditioal operator and said propositions are not implications.

So I'm unclear why you think "the false on both sides /condition/ is relevant when a biconditional functions as a definition.

>>So now we turn to the semantics. A definition of a square must include all squares and it must /exclude/ all non-squares. That's what the 'iff' does.<<

Agreed

>>In particular, the property of the biconditional being true when both LHS and RHS are false can be understood to mean that it is the case that if you encounter a shape and it is false that it is a square,..<<

Equivalent to is true that the figure is not a square.

>>then it is false that all four sides are congruent and all four angles are congruent.<<
Equivalent to to it is true that not (then it is false that all four sides are congruent and all four angles are congruent).
>> If it is false that all four sides are congruent and all four angles are congruent, then it is false that it is a square.<<
Agreed

>>But the /definition/ is true<<

Indeed. Unfortunately it is a true definition of all figures that are not squares.

>>I hope that clears up the issue<<

Well not really since your argument gives to mutually exclusive sets, all figures that are not squares and all figures that are squares.

Hi Bx4,

>>What some call pedantry others call precision. <<

I'm betting you're one of the others...

>>But also inaccurate. <<

The meaning seemed perfectly clear to me, I'm sorry it wasn't to you.

>>I disagree with your interpretation of 'forward' and backward' as (necessarily?) requiring recourse the the biconditional expansion since I don't think:

'A biconditional statement can be either true or false. To be true, //BOTH// the conditional statement and its converse must be true. This means that a true biconditional statement is true both "forward" and "backward". //All definitions// can be written as //true// biconditionals' [emphases added]

says this. <<

It does though. The conditional statement in the expanded form of p <-> q.r is p -> q.r. This supports my interpretation, and a similar example is given in your linked slideshow.

>>A simpler interpretation is that 'forward' means<<

Some might say 'simpler', others...

>>p<=>q.r

and 'backward' means

q.r<=>p <<

Switching the LHS and RHS doesn't change the direction since both are bidirectional, so no 'forward' or 'backward' occurs without the correct concept of conditional statement: p -> q.r (forwards). All that changes in your switch is direction of reading, which is a lexical rather than logical phenomenon, and a trivial one at that.

>>Indeed though I later changed my position when I said that I ha realised that I disd not need the biconditional expansion to support my argument and was happy to use the unexpanded biconditional where the terms 'conditional' (LHS) and converse (RHS) do not have the same meaning as your usage above. <<

The only person who I have ever seen interpret 'conditional' as synonymous with LHS is you. I think this is an error on your part. If you disagree, perhaps you could give some examples of this usage in the literature or alternatively an argument of your own to support this interpretation?

>>In the biconditonal 'conditional' simply means the proposition or propositions to the left hand of the biconditioal operator and 'converse' simply means the proposition or propositions to the right the biconditioal operator and said propositions are not implications. <<
I think this is the source of our disagreement and I pointed out that this was an error. It makes no sense to identify 'the conditional statement' with the LHS. Your own quote and linked example uses the term 'conditional statement' in the same sense as I have, namely p -> q.r and the converse is q.r -> p. I have pointed to the evidence for this and it is clear. I can't see what else I can do to facilitate progress here.

>>So I'm unclear why you think "the false on both sides /condition/ is relevant when a biconditional functions as a definition. <<

And so you will remain until you sort out your terminology, otherwise you will continue to think the LHS and RHS have to be true for a definitional biconditional. That's just wrong, rather the conditional statement and its converse in the equivalent expanded form must be true. That's what it says in your own link and quote from #143.

>>Indeed. Unfortunately it is a true definition of all figures that are not squares.<<
No I think you've missed the point. It is a true definition of square in virtue of what it includes (all squares) and, as I illustrated, what it exludes (all non-squares).

>>Well not really since your argument gives to mutually exclusive sets, all figures that are not squares and all figures that are squares.<<
See above.

ttfn

hi psi

>>See above<<

Ok to cut to the chase:

Given the biconditional:

(a)p<=>q.r


I am quite happy,to:

(1) Use the terminology LHS and RHS to refer to terms on either side of the biconditional operator.

(2) Avoid recourse to the biconditional expansion as it is irrelevant to my argument.

(3) use the biconditional truth tables which only involves the variables p and q.r and the bicondtional operator

(4) Use the specific biconditional definition:

(b)'A figure is a square if and only if all four sides are congruent and all four angles are congruent.'


so that you can show how

(b.1)It is true that A figure is a square if and only if it is true all four sides are congruent and all four angles are congruent.

and

(b.2)It is false that A figure is a square if and only if it is false all four sides are congruent and all four angles are congruent.

Are both biconditionals which define the set of all figures that are squares rather than (b.1) being the biconditional which defines the set of all figures that are squares and (b.2) being the biconditional which defines the set of all figures that are not squares.

bs

Hi Bx4,

>>Ok to cut to the chase:<<

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

>>Given the biconditional:

(a)p<=>q.r


I am quite happy,to:

(1) Use the terminology LHS and RHS to refer to terms on either side of the biconditional operator.

(2) Avoid recourse to the biconditional expansion as it is irrelevant to my argument.

(3) use the biconditional truth tables which only involves the variables p and q.r and the bicondtional operator

(4) Use the specific biconditional definition:

(b)'A figure is a square if and only if all four sides are congruent and all four angles are congruent.'<<

That's fine, but as I said before, (2) deprives you of the very thing you are relying on, namely that a true definitional biconditional cannot use the fourth row of the truth table, where both the LHS and the RHS are false. Unless you can find some other support for this notion? The argument you have offered below doesn't seem relevant.

>>so that you can show how

(b.1)It is true that A figure is a square if and only if it is true all four sides are congruent and all four angles are congruent.

and

(b.2)It is false that A figure is a square if and only if it is false all four sides are congruent and all four angles are congruent.

Are both biconditionals which define the set of all figures that are squares rather than (b.1) being the biconditional which defines the set of all figures that are squares and (b.2) being the biconditional which defines the set of all figures that are not squares.

bs <<

How is this relevant? What we have is the biconditional (b). If p or q or r happen to be false this does not change the biconditional b itself into b.2, whereas it is nonsensical to incorporate 'is true' as in b1 into the wording of the biconditional. Recall Ramsay's Ladder?

Have you looked through the slides in your link? The interpretation of directions 'forward' and 'backward' and the meaning of 'conditional' and 'converse' are clear. They are as I have described them.

Perhaps you see a meaningful destinction, but if you give precise conditions such that iff they are met, S is a square, what changes if you frame them in the negative? if you say S is not a square iff they are not met, that's equivalent.

Just to be clear, suppose we encounter a shape S and we want to use b to see whether it is a square. We have:

(b)'A figure is a square if and only if all four sides are congruent and all four angles are congruent.'

Suppose all four angles in S are not congruent. Then b is true only if we say that it is false that S is a square. That's because, if we let p = S is a square, q = S has all four sides congruent and r = S has all four angles congruent, we can write b as:

p <=> q.r

p = F
q = T (let's say S is a rhombus)
r = F
q.r = F (law of conjunction)

b = T since F<=>F = T

Note however, that at no stage did we need to change b or invoke something akin to b.2
( ¬p <=> ¬(q.r)).

We do not need to incorporate the negative into the syntax of the proposition when we consider particular cases in which a propositional variable takes the value F.

I hope that makes sense.

ttfn