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instantiation of EA formulas on finite domains
Can someone tell me if this is correct?

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instantiation of EA formulas on finite domains

moonlight
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Posted 11/06/07 - 09:27 AM: Subject: instantiation of EA formulas on finite domains	#1
Given a formula F, whereby F = (Ex)(Ay)U(x,y) and where the functions in U range over a domain D, whereby \|D\| = 3. The elements of D are denoted a, b, c. Can the formula F be instantiated as follows? F = (Ex)(Ay)U(x,y) (x,y range over D) F = (Ay) [U(a,y) v U(b,y) v U(c,y)] (y ranges over D) F = [U(a,a) ^ U(a,b) ^ U(a,c)] v [U(b,a) ^ U(b,b) ^ U(b,c)] v [U(c,a) ^ U(c,b) ^ U(c,c)] Is this correct? Edited by moonlight on 11/06/07 - 09:34 AM
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Posted 11/06/07 - 09:30 AM:	#2
error: double post
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HamishMacSporran
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Posted 11/07/07 - 01:55 AM:	#3
moonlight wrote: Given a formula F, whereby F = (Ex)(Ay)U(x,y) and where the functions in U range over a domain D, whereby \|D\| = 3. The elements of D are denoted a, b, c. Can the formula F be instantiated as follows? F = (Ex)(Ay)U(x,y) (x,y range over D) F = (Ay) [U(a,y) v U(b,y) v U(c,y)] (y ranges over D) F = [U(a,a) ^ U(a,b) ^ U(a,c)] v [U(b,a) ^ U(b,b) ^ U(b,c)] v [U(c,a) ^ U(c,b) ^ U(c,c)] Is this correct? No, F = (Ex)(Ay)U(x,y) says that there is some x for which all y have the property U(x,y). So, the expansion in terms of disjunction should be: F = (Ay) [U(a,y)] v (Ay) [U(b,y)] v (Ay) [U(c,y)] Your final expansion is correct though, but isn't an expansion of the formula above it.

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Posted 11/13/07 - 05:52 PM:	#4
HamishMacSporran wrote: No, F = (Ex)(Ay)U(x,y) says that there is some x for which all y have the property U(x,y). So, the expansion in terms of disjunction should be: F = (Ay) [U(a,y)] v (Ay) [U(b,y)] v (Ay) [U(c,y)] Your final expansion is correct though, but isn't an expansion of the formula above it. I'm sorry I'm answering this late. But thanks a lot for your response. I really appreciate as I was confused on this issue. I thought that (Ex)(Ay)U(x,y) could be read as (Ex)[(Ay)U(x,y)] and from there inferred that we would instantiate the (Ay) first. So thanks for clarifying that! Cordially, moonlight.
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Posted 03/18/08 - 09:27 PM:	#5
HamishMacSporran wrote: No, F = (Ex)(Ay)U(x,y) says that there is some x for which all y have the property U(x,y). So, the expansion in terms of disjunction should be: F = (Ay) [U(a,y)] v (Ay) [U(b,y)] v (Ay) [U(c,y)] Your final expansion is correct though, but isn't an expansion of the formula above it. Hi, I know it's been a while, but I just noticed something in your answer and I think it needs to be corrected. Your expansion is actually false, and the original one I presented was correct. This is because by having Ay. 3 times you are implying that each y is a different variable. The expansion you recommended is equivalent in prenex normal form to a formula with a prefix containing 3 variables. Cordially, moonlight.
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Posted 03/19/08 - 10:35 AM:	#6
ExAy Uxy Unpack 'Ex': Ay Uay v Ay Uby v Ay Ucy Unpack each of the 'Ay': (Uaa & Uab & Uac) v (Uba & Ubb & Ubc) v (Uca & Ucb & Ucc)

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Posted 03/19/08 - 11:17 AM:	#7
Hi MoeBlee, MoeBlee wrote: ExAy Uxy Unpack 'Ex': (1) Ay Uay v Ay Uby v Ay Ucy This is the problem I'm having here. When this formula (1) is put in prenex normal form it yields: (2) Ay.Ay.Ay. Uay v Uby v Ucy. Which then requires alpha renaming to: (3) Ax.Ay.Az. Uax v Uby v Ucz. After all, when given formula (1) nothing says that each of the 'Ay.' is binding the same variable. Don't you think that this is an issue here? Cordially, moonlight.
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Posted 03/19/08 - 12:17 PM:	#8
Theorem: Ayxz(Uay v Ubx v Ucz) <-> Ay Uay v Ay Uby v Ay Ucy.

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Posted 03/19/08 - 12:18 PM:	#9
Hi Moonlight, I’m not sure about Hamish’s interpretation implying 3 variables. If a formula repeats a sign I think we should adopt the convention that the sign means the same thing throughout the formula unless such a reading cannot be sustained. The real problem here I feel is that the formula is also ambiguous in more ways than one. I was wondering about the difference between: Let U = uncle, then F could be interpreted as ‘some x is the uncle of all y‘. Here x is the uncle of all y meaning each individual member of y, and the whole class represented by y grouped together.They could be triplets! Either way does not seem to matter. But there is another ambiguous interpretation: Let U` = sound, then F could be interpreted ‘some x is the sound of all y; Is x the sound of each individual y, or is it the sound of all the y added together? The sum noise of the crowd compared to the sound of an individual taken in isolation. I don’t think the original formula (Ex)(Ay)U(x,y) specifies which. But the interpretative stance we take will define the resulting series. Okay. I’m a bit reluctant to write what follows because I’d really like to write things completely differently in light of our other conversation and the system we have been discussing there. However I would like to draw out the ambiguities at play, and so I put these next thoughts forward for heuristics reasons only. Anyhow…. If U = the individual sound of any y, then I might write: 1. [U(a,a) v U(a,b) v U(a,c)] v [U(b,a) v U(b,b) v U(b,c)] v [U(c,a) v U(c,b) v U(c,c)] If U = the sound of all y grouped together, which I think is the actual implication of what Hamish’s tries to write with F = (Ay) [U(a,y)] v (Ay) [U(b,y)] v (Ay) [U(c,y], I might write: 2. [U(a,(a ^ b ^ c))] v [U(b,(a ^ b ^ c))] v [U(c,(a ^ b ^ c))] If U = uncle of every y, as per moonlight I might write: 3. [U(a,a) ^ U(a,b) ^ U(a,c)] v [U(b,a) ^ U(b,b) ^ U(b,c)] v [U(c,a) ^ U(c,b) ^ U(c,c)] As for (Ex)(Ay)U(x,y) I think it can be interpreted in all three ways...I think.

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Posted 03/19/08 - 12:26 PM:	#10
Also, I don't think your (2) is PNF for (1). Rather, (3), which you call a "renaming", is PNF for (1).

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Posted 03/19/08 - 12:38 PM:	#11
Furrowed Brow 2 wrote: The real problem here I feel is that the formula is also ambiguous in more ways than one. No, it's not ambiguous. The unique readability theorem for formulas ensures that no well formed formula is ambiguous given an interpretation. Furrowed Brow 2 wrote: Let U = uncle, then F could be interpreted as ‘some x is the uncle of all y‘. Here x is the uncle of all y meaning each individual member of y, NO, that is not what it would mean at all. Whatever ambiguity you find there comes from English and not from the formula. And your ambiguity is just a silly one. When we "say x is, for all y, an uncle of y" we surely don't mean that y is set of nephews and nieces. That would be being silly in twisting perfectly sensical English. Furrowed Brow 2 wrote: Let U` = sound, then F could be interpreted ‘some x is the sound of all yIs x the sound of each individual y, or is it the sound of all the y added together? Each individual. Your problems with that thereby resolved. There is nothing in either the formula or even the English interpretation that says anything about "adding together".

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Posted 03/19/08 - 12:54 PM:	#12
MoeBlee wrote: Theorem: Ayxz(Uay v Ubx v Ucz) <-> Ay Uay v Ay Uby v Ay Ucy. Indeed. How silly of me not to have noticed that. Regarding the PNF of 1, you're also right: 3 is the PNF. Alpha renaming must be carried out before the sentence is put in any special normal form, so as to avoid ambiguities. Thanks again MoeBlee for your input. Cordially, moonlight.
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Posted 03/19/08 - 01:09 PM:	#13
Hi Furrowed Brow, Furrowed Brow 2 wrote: I’m not sure about Hamish’s interpretation implying 3 variables. If a formula repeats a sign I think we should adopt the convention that the sign means the same thing throughout the formula unless such a reading cannot be sustained. Yes, but there is already a standard convention. When a variable is inside the scope of two (or more) quantifiers it is bound by the most local one. So the 3rd 'y' is bound by the 3rd 'A' in sentence (3) for example. Moreover when you have distinct variables which share the same name in a formula, then you rename them so as to avoid any ambiguities: the technique is called alpha-renaming and was originally developped by Church for the lambda calculus. In this case, it turns out that it doesn't matter whether the variables are renamed or not, both formulas are equivalent. I'm not sure this holds for all type of formulas though. Furrowed Brow 2 wrote: Let U = uncle, then F could be interpreted as ‘some x is the uncle of all y‘. Here x is the uncle of all y meaning each individual member of y, and the whole class represented by y grouped together.They could be triplets! Either way does not seem to matter. But there is another ambiguous interpretation: Let U` = sound, then F could be interpreted ‘some x is the sound of all y; Is x the sound of each individual y, or is it the sound of all the y added together? The sum noise of the crowd compared to the sound of an individual taken in isolation. I don’t think the original formula (Ex)(Ay)U(x,y) specifies which. But the interpretative stance we take will define the resulting series. Well if you use the standard semantics, then there is no ambiguity. The formula states that on some domain D, which is a set, containing elements a_1 to a_n, then there is some element 'a_i' of D, such that every ordered pair (a_i, a_1) up to (a_i, a_n) is in the extension of the set U, which is itself a subset of DxD. It doesn't matter then what the domain D contains, or what the relation U is meant to denote. Obviously this formula could be true for some interpretation, and false for others. So it isn't a theorem. Cordially, moonlight. P.S: Sorry for not having answered back yet on the other topic. I'll come back to it as soon as I can. I just need to have a serious amount of time (couple of hours at least) to re-examine your system and give you feedback.
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Posted 03/19/08 - 01:34 PM:	#14
Moeblee wrote: No, it's not ambiguous. The unique readability theorem for formulas ensures that no well formed formula is ambiguous given an interpretation. Okay. And I’ve think there’s three interpretations…one if we count it your way. Moeblee wrote: Each individual. Your problems with that thereby resolved. There is nothing in either the formula or even the English interpretation that says anything about "adding together". I think you’re wrong about the English interpretation. Natural language is full of ambiguities. And these are logical ambiguities not silly. We need a formula for each interpretation. Help me out here then. x is the noise made by everyone. Two possible interpretations…yes? Everyone individual might be shouting “goal!” and that will be the noise of one, or everyone might sound like a roar. The English sentence does not specify which interpretation. Both are possible…which one do we give? What formula do we use to distinguish the noise made by any one, and the noise made by every one (the noise of the crowd)? If quantified formula are to only be used to express individual relationships to a predicate, how do we write formula that relates the group to a predicate like the roar made by a group of individuals?

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Posted 03/19/08 - 02:39 PM:	#15
Furrowed Brow 2 wrote: What formula do we use to distinguish the noise made by any one, and the noise made by every one (the noise of the crowd)? Well, if R(x,y) <=> x "makes noise" y, then by the standard interpretation: Ax.Ey. R(x,y) <=> Everyone makes some noise. Not necessarily the same noise. Ey.Ax. R(x,y) <=> At least one noise is made by everyone. This is your crowd. Cordially, moonlight.
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Posted 03/19/08 - 02:52 PM:	#16
Furrowed Brow 2 wrote: Natural language is full of ambiguities. And these are logical ambiguities not silly. I'm NOT disputing that natural language has ambiguities. But the formulas of the predicate calculus aren't ambiguous in the very sense I mentioned earlier. The ambiguities are ones of natural language and they are not "logical ambiguities". And what was silly was your attempt to ludicrously stretch the possible range of ambiguity of a particular interpretation and thus, doubly illogically, to argue that there is an ambiguity in the logic formula itself. On the other hand, that there are natural language expressions that don't need to be so ludicrously stretched to evince ambiguity is not contested and is quite well known to logicians. Don't confuse the direction of translation: Yes, translating from a natural language expression to a logical formula may well be an effort mired by ambiguity. But translating a logical formula into a natural language expression is reasonaly unambiguous GIVEN that the context of translation (the interpretation of the formal language) is reasonably unambiguous. So, in the instance where 'U' is taken to stand for 'is an uncle of', and 'is an uncle of' is understood in some reasonably unambiguous way, translating the formula 'ExAy Uxy' is virtually unambiguous. It clearly says, "There is an individual one who is an uncle of everybody". And in a domain of three individuals (say, three people), the meaning is combinatorically exact, though one might wince at countenancing the inbreeding that would have to occur to make the sentence true. Edited by MoeBlee on 03/19/08 - 03:09 PM

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Posted 03/19/08 - 03:25 PM:	#17
MoeBlee wrote: And in a domain of three individuals (say, three people), the meaning is combinatorically exact, though one might wince at countenancing the inbreeding that would have to occur to make the sentence true. When mathematics goes immoral moonlight.
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Posted 03/20/08 - 04:27 PM:	#18
Hi moonlight, Chill. No problem about feedback. Frankly I appreciate your effort. Anyhow apologies for derailing your thread with messy semantics. moonlight wrote: Furrowed Brow 2 wrote: What formula do we use to distinguish the noise made by any one, and the noise made by every one (the noise of the crowd)? Well, if R(x,y) <=> x "makes noise" y, then by the standard interpretation: Ax.Ey. R(x,y) <=> Everyone makes some noise. Not necessarily the same noise. Ey.Ax. R(x,y) <=> At least one noise is made by everyone. This is your crowd. Thanks for that. Appreciate the patience. But I’m not sure this clears things up for me. For your second example we could also offer the interpretation: “At least one white lie is told by everyone” - meaning everyone tells the same white lie. The lie does not necessarily grow for being repeated. My copy of Lemmon interprets (Ey)(x)Pxy as “someone has everyone as a parent”. Okay that is false, or we’re allowing for strange DNA experiments, but an extraordinary number of multiple parents does not entail an increase in any of the properties of the someone other than an increase in the number of parents. If we just follow standard interpretations then as a matter of form that would be like each person in the crowd repeating the same word “Goal!” or repeating the same noise - but their efforts have no accumulative properties. For a limited domain of three then each could be repeating the same noise I.e. “goal!”, “goal!” and “goal”. However I’m thinking of the group/crowd noise “GOAL!!!””. A formula that expresses the point that each member of a crowd makes the same noise indeed expresses the characteristics property of each member of the group, but each individual noise considered individually does not raise the roof, does not carry for several miles, does not intimidate the opposition, does not spur on your team, does not engender group identity etc. So I do not think Ey.Ax. R(x,y) is the right formula to express an accumulative R = NOISE!!! Which may have its own group characteristic properties. Okay I‘ve looked at this again. Let U = noise! F = [U(a,a) ^ U(a,b) ^ U(a,c)] v [U(b,a) ^ U(b,b) ^ U(b,c)] v [U(c,a) ^ U(c,b) ^ U(c,c)] If we allow that this is the correct interpretation (deep down I don’t really) then what for U = NOISE!!! And the NOISE!!! made by everyone. I guess we could talk about the roar of the crowd, but the crowd is made up of individuals. So we are still talking about a predicate that applies to everyone.

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Posted 03/20/08 - 04:27 PM:	#19
Hi moonlight, Chill. No problem about feedback. Frankly I appreciate your effort. Anyhow apologies for derailing your thread with messy semantics. moonlight wrote: Furrowed Brow 2 wrote: What formula do we use to distinguish the noise made by any one, and the noise made by every one (the noise of the crowd)? Well, if R(x,y) <=> x "makes noise" y, then by the standard interpretation: Ax.Ey. R(x,y) <=> Everyone makes some noise. Not necessarily the same noise. Ey.Ax. R(x,y) <=> At least one noise is made by everyone. This is your crowd. Thanks for that. Appreciate the patience. But I’m not sure this clears things up for me. For your second example we could also offer the interpretation: “At least one white lie is told by everyone” - meaning everyone tells the same white lie. The lie does not necessarily grow for being repeated. My copy of Lemmon interprets (Ey)(x)Pxy as “someone has everyone as a parent”. Okay that is false, or we’re allowing for strange DNA experiments, but an extraordinary number of multiple parents does not entail an increase in any of the properties of the someone other than an increase in the number of parents. If we just follow standard interpretations then as a matter of form that would be like each person in the crowd repeating the same word “Goal!” or repeating the same noise - but their efforts have no accumulative properties. For a limited domain of three then each could be repeating the same noise I.e. “goal!”, “goal!” and “goal”. However I’m thinking of the group/crowd noise “GOAL!!!””. A formula that expresses the point that each member of a crowd makes the same noise indeed expresses the characteristics property of each member of the group, but each individual noise considered individually does not raise the roof, does not carry for several miles, does not intimidate the opposition, does not spur on your team, does not engender group identity etc. So I do not think Ey.Ax. R(x,y) is the right formula to express an accumulative R = NOISE!!! Which may have its own group characteristic properties. Okay I‘ve looked at this again. Let U = noise! F = [U(a,a) ^ U(a,b) ^ U(a,c)] v [U(b,a) ^ U(b,b) ^ U(b,c)] v [U(c,a) ^ U(c,b) ^ U(c,c)] If we allow that this is the correct interpretation (deep down I don’t really) then what for U = NOISE!!! And the NOISE!!! made by everyone. I guess we could talk about the roar of the crowd, but the crowd is made up of individuals. So we are still talking about a predicate that applies to everyone.

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Posted 03/20/08 - 04:27 PM:	#20
Hi moonlight, Chill. No problem about feedback. Frankly I appreciate your effort. Anyhow apologies for derailing your thread with messy semantics. moonlight wrote: Furrowed Brow 2 wrote: What formula do we use to distinguish the noise made by any one, and the noise made by every one (the noise of the crowd)? Well, if R(x,y) <=> x "makes noise" y, then by the standard interpretation: Ax.Ey. R(x,y) <=> Everyone makes some noise. Not necessarily the same noise. Ey.Ax. R(x,y) <=> At least one noise is made by everyone. This is your crowd. Thanks for that. Appreciate the patience. But I’m not sure this clears things up for me. For your second example we could also offer the interpretation: “At least one white lie is told by everyone” - meaning everyone tells the same white lie. The lie does not necessarily grow for being repeated. My copy of Lemmon interprets (Ey)(x)Pxy as “someone has everyone as a parent”. Okay that is false, or we’re allowing for strange DNA experiments, but an extraordinary number of multiple parents does not entail an increase in any of the properties of the someone other than an increase in the number of parents. If we just follow standard interpretations then as a matter of form that would be like each person in the crowd repeating the same word “Goal!” or repeating the same noise - but their efforts have no accumulative properties. For a limited domain of three then each could be repeating the same noise I.e. “goal!”, “goal!” and “goal”. However I’m thinking of the group/crowd noise “GOAL!!!””. A formula that expresses the point that each member of a crowd makes the same noise indeed expresses the characteristics property of each member of the group, but each individual noise considered individually does not raise the roof, does not carry for several miles, does not intimidate the opposition, does not spur on your team, does not engender group identity etc. So I do not think Ey.Ax. R(x,y) is the right formula to express an accumulative R = NOISE!!! Which may have its own group characteristic properties. Okay I‘ve looked at this again. Let U = noise! F = [U(a,a) ^ U(a,b) ^ U(a,c)] v [U(b,a) ^ U(b,b) ^ U(b,c)] v [U(c,a) ^ U(c,b) ^ U(c,c)] If we allow that this is the correct interpretation (deep down I don’t really) then what for U = NOISE!!! And the NOISE!!! made by everyone. I guess we could talk about the roar of the crowd, but the crowd is made up of individuals. So we are still talking about a predicate that applies to everyone.

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Posted 03/20/08 - 04:29 PM:	#21
Hi Moeblee Moeblee wrote: I'm NOT disputing that natural language has ambiguities. Point taken. Moeblee wrote: But the formulas of the predicate calculus aren't ambiguous in the very sense I mentioned earlier. The ambiguities are ones of natural language and they are not "logical ambiguities". Again point taken. Moeblee wrote: And what was silly was your attempt to ludicrously stretch the possible range of ambiguity of a particular interpretation and thus, doubly illogically, to argue that there is an ambiguity in the logic formula itself. Moeblee I accept the fault lies with me. Thank you for pointing out where the line of credibility lies. I’ll try, but I can’t promise not to cross it again. Moeblee wrote: On the other hand, that there are natural language expressions that don't need to be so ludicrously stretched to evince ambiguity is not contested and is quite well known to logicians. Don't confuse the direction of translation: Yes, translating from a natural language expression to a logical formula may well be an effort mired by ambiguity. But translating a logical formula into a natural language expression is reasonably unambiguous GIVEN that the context of translation (the interpretation of the formal language) is reasonably unambiguous. Again point taken. Let the logical syntax and vocabulary of the formal language dictate the form and limits of the natural language interpretation - and you can’t go far wrong. OK with you. - Small worry at the back of my brain: if the predicate calculus is a general predicate logic it will be able to express every logical relationship involving predicates without ambiguity….none left over…yes? So despite the ambiguities and difficulties of natural language, as was the hopes of original project set in train by Frege, every semantic interpretation has a corresponding predicate formula. Moeblee wrote: So, in the instance where 'U' is taken to stand for 'is an uncle of', and 'is an uncle of' is understood in some reasonably unambiguous way, translating the formula 'ExAy Uxy' is virtually unambiguous. It clearly says, "There is an individual one who is an uncle of everybody". And in a domain of three individuals (say, three people), the meaning is combinatorically exact, though one might wince at countenancing the inbreeding that would have to occur to make the sentence true. Okay. I’ll not hang out with uncle and that side of the family. But things are still a bit blurry for me. 'ExAy Uxy' - "There is an individual one who is an uncle of everybody". Ex.Ay.Mxy - “There is an x that is the mass of every y”. Following your example the formal rules that apply to Ex.Ay.Mxy clearly mean the individual mass that applies to each y. So all y have the same mass, and we do not add masses I.e. Ma, Ma, Ma and not Ma + Ma + Ma. Consequently we can’t use that same formula for: Formula? - “There is an x that is the mass of every y”. Meaning: The value that is the mass of every y I.e. Ma + Ma + Ma and not Ma, Ma, Ma. Once we have tied down an unambiguous formula that expresses this interpretation and not the former then everything will be cleared up nicely. I think moonlight also has a related problem of crowd control you might point me right on. If all this is old ground already well trod by logicians then if you can supply a link I’d be grateful and I would not ask you to respond in depth.

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Posted 03/21/08 - 09:23 AM:	#22
Furrowed Brow 2 wrote: Again point taken. Let the logical syntax and vocabulary of the formal language dictate the form and limits of the natural language interpretation - and you can’t go far wrong. I wouldn't go that far. It could be that the logic system is not adequate to capture the complexity of the natural language expression for some purpose. The predicate calculus is not designed to be able to capture the meaning and deductive relations among all possible natural language expressions. Rather, the first order predicate calculus is mainly expected only to capture the (extensional) meanings and deductive relations among natrual language expressions of ordinary mathematics. And what I'm saying about lack of ambiguity is this: (1) If the natural language expression is ambiguous, then when we express it by a predicate calculus formula, we have to choose just one of the meanings of the natural language expression. (2) The main point I made is that given the predicate calculus formula FIRST and THEN a a reasonalby unambiguous context of interpretation, the resulting interpretation will be reasonably unambiguous. (Moreover, if the context of interpretation is mathematical, we can make the interpretation PERFECTLY unambiguous (though not necessarily computable).) Furrowed Brow 2 wrote: if the predicate calculus is a general predicate logic it will be able to express every logical relationship involving predicates without ambiguity….none left over…yes? I don't understand your question. First, the expressive power of predicate calculus IS limited. Second, I don't know what you mean by "left over predicates". Furrowed Brow 2 wrote: 'ExAy Uxy' - "There is an individual one who is an uncle of everybody". Right. Furrowed Brow 2 wrote: Ex.Ay.Mxy - “There is an x that is the mass of every y”. Right. Furrowed Brow 2 wrote: Following your example the formal rules that apply to Ex.Ay.Mxy clearly mean the individual mass that applies to each y. So all y have the same mass, Right. Furrowed Brow 2 wrote: and we do not add masses I.e. Ma, Ma, Ma and not Ma + Ma + Ma. No, there's nothing to stop us from adding masses. But in this case, the operation of adding masses will be idempotent. There is only one mass, so adding itself to itself results in itself. Of course, though, where 'mass' is understood in the ordinary sense or in physics, 'ExAyMxy' is false and addition of masses is not idempotent. The formula is false where 'M' is interpreted as 'the mass of', but still the correct interpretation is "all objects have the same mass as one another".

moonlight
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Posted 03/21/08 - 04:30 PM:	#23
Hi Furrowed Brow, Furrowed Brow 2 wrote: The lie does not necessarily grow for being repeated. My copy of Lemmon interprets (Ey)(x)Pxy as “someone has everyone as a parent”. Okay that is false, or we’re allowing for strange DNA experiments, but an extraordinary number of multiple parents does not entail an increase in any of the properties of the someone other than an increase in the number of parents. If we just follow standard interpretations then as a matter of form that would be like each person in the crowd repeating the same word “Goal!” or repeating the same noise - but their efforts have no accumulative properties. For a limited domain of three then each could be repeating the same noise I.e. “goal!”, “goal!” and “goal”. However I’m thinking of the group/crowd noise “GOAL!!!””. I see what you mean. I think that you are having an issue with the order of the quantification here. Basically, what you're saying is that first order quantification is not enough for you. There is a famous example about this: "the men lifted the Piano together". Formalizing this sentence is a bit problematic with first order logic. Obviously with 1st order quantifiers you could say that each man lifted the piano, but it does not really express the point: that they lifted it *together. So what you need is further quantification on the whole set of guys to express that it is the set of guys, not just each guy one by one, who lifted the piano. But semantically a set, is a predicate. So in effect you would be quantifying over a predicate and this is second order quantification as opposed to first order quantification where you only quantify over individuals. The philosopher Boolos argued for an existential monadic second order quantification to solve this issue in the 70's and 80's. See his famous paper: to be is to be the value of a variable*. He called this "Plural Quantification". Quine strongly argued against this because he claimed it meant an ontological commitment to a set, which is not supposed to be an entity. Well, I'm no expert on the issue, but basically this is what I wanted to say: you are contemplating second order quantification without reliazing it perhaps, since you want to say something about the crowd as a whole, not just each individual within it. You seem to want to consider the crowd itself as an object. That would have made Quine real mad. Cordially, moonlight.
_____________________ All are lunatics, but he who can analyze his delusion is called a philosopher. - Ambrose Bierce -

Furrowed Brow 2
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Posted 03/22/08 - 06:42 AM:

#24

Hi moonlight

Ah ha…I think we are moving into interesting territory.

moonlight wrote:
There is a famous example about this: "the men lifted the Piano together".

Excellent example. Better than mine. Makes the point clearly and succinctly. I’m going to start using this example. Thank you.

moonlight wrote:
So what you need is further quantification on the whole set of guys to express that it is the set of guys, not just each guy one by one, who lifted the piano.

Yes, but I think we need to be careful as to what you think I think than entails.

moonlight wrote:
But semantically a set, is a predicate.

I’d say a set is represented by a collection of signs that are the values of the variable sign, and as such are each related to the same predicate sign in the same way.

moonlight wrote:
So in effect you would be quantifying over a predicate and this is second order quantification as opposed to first order quantification where you only quantify over individuals.

I think I’ve mentioned in our other discussion that I do not believe quantifying over predicates makes any sense. We can only quantify over the values of the variable. So I don’t think the answer to this problem is second order. In fact I’m sure it can’t be.

I see this is a problem of how we represent the set. Let L = lives,

With the two series interpreted as is standard then the first series should comes out as a universally quantified group, and the second existentially quantified. But I’m approaching at my own idiosyncratic tangent. Two version of a quantified formula can be constructed (I mean constructed with the modified tables).

Let the square brackets [ ] represent a modified truth table. The variable placed outside the brackets, let + = signifies, and the minus - = fails to signify, let the curl ~ = deny any agreement in how signs signify. For example -Lx [-a ^ -b ^ -c] is another way of representing the modified truth table (F-FFFFFFFFFFFF-T)(Lx,a,b,c).Let L = lifts. So:

1/ +Lx [ +a ^ +b ^ + c] and its negation = -Lx [-a v -b v -c] give:

(Ax).Lx = La ^ Lb ^ Lc or La v Lb v Lc
2/ +Lx [ +a v +b v + c] and its negation = -Lx [-a ^ -b ^ -c] give

(Ax).Lx = La ^ Lb ^ Lc or La v Lb v Lc
3/ +Lx [ -a ^ -b ^ -c] and its negation = -Lx [+a v +b v +c] give:

(Ax).~Lx = ~La ^ ~Lb ^ ~Lc or ~La v ~Lb v ~Lc
4/ -Lx [ +a v +b v + c] and its negation = +Lx [-a ^ -b ^ -c] give

(Ax).~Lx = ~La v ~Lb v ~Lc or ~La ^ ~Lb ^ ~Lc
We then get two versions of the universal quantifier that signify their objects

The negative propostion is intresting because it neatly covers all the semantic bases.

The two different series able to express the different ways the signs form up in a series. (Things are a bit more involved and the above presentation needs tweaking - but I’m keeping it simple).

I think this approach resolves a lot of problems and in the final fully tweaked version causes none. For instance we can quantifying over empty domains. And this account may go someway to explaining why I see ambiguity in predicate formula in the prenex form. It also explains my preoccupation with the ambiguities of natural language sentences. We do not have to move into second order logic to resolve the piano lifting problem. We just need to readdress predicate logic with truth table constructions.

I'll let you digest this first. We might then take a look at how I'd write the formula that you started with.

I’m tempted to say more but I can hear Moeblee’s foot falls running down the corridor to come put out the fire

FB

Edited by Furrowed Brow 2 on 03/22/08 - 07:47 AM

moonlight
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Posted 03/22/08 - 03:21 PM:	#25
Hi FB, Well, I was talking in my case about the usual system of logic. It may very well be that you have a different system that doesn't need such higher order quantification, although I can't tell: I find your system really hard to learn. I just can't make sense of how the + "signifies" and - "fails to signify" and the ~ are used and what exactly they mean. I'd really need more to time to learn about it. I know this must sound frustrating to you. Perhaps if you had a simpler introduction with comparisons with the usual notation and better examples on your site it would help? Also, the quantification over empty domains is an audacious claim. In an empty domain a sentence like: Ax. P(x) would be true, and its negation would also be true. That leads to inconsistency in the system of logic. It sounds bizzare but, I'm keeping a mental note to check out your system and try to see the correspondence at some later time. This is research in itself. I do recommend though that you look at Boolos' and Quine's works and opposing views on this type of examples too however. You may find it interesting. You may agree with Quine who doesn't think second order quantification for such sentences is a good idea. Here's another example like the piano one that you may like: "there are some critics who admire only one another" (a.k.a the infamous Geach-Kaplan sentence). Check out "plural quantification" on the Stanford Encyclopedia of Philosophy for explanations on the sentence's implications. http://plato.stanford.edu/entries/plural-quant/ Cordially, moonlight.
_____________________ All are lunatics, but he who can analyze his delusion is called a philosopher. - Ambrose Bierce -

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