View source for Is ⟨me'i⟩ importing?


Why is me'i importing?  Simply because "less than zero" makes no sense for discrete objects? -- RobinLeePowell

* Because '''me'i''' is take to be equivalent to '''naku ro''' (''not all''). Since it has been agreed that '''ro''' does not have existential import, its negation must have existential import, and ''me'i'' is its negation. (We could consider that '''me'i''' is not strictly a negation of '''ro''', but no one has argued for that so far this round, and traditionally in both logic and natural languages there is no separate quantifier for ''not all'', just ''all'' plus negation.) -- AdamRaizen
* If there are 9 broda, then ro broda = 9 broda, & so I don't see why '''me'i (ro) broda''' can't = '''0 broda'''. As is said in the main text above, it hasn't been agreed that '''me'i''' is the right cmavo for the job. And even if it turns out that it is, we haven't yet got the reasons straight. --[[User:And Rosta|And Rosta]]
** Ok '''me'i''' is definitely a bad choice.  I should've said this when And asked for comments:  the quantifier is supposed to mean "Some are not", i.e. '''su'o da naku ...''' (this is deductively equivalent to the '''naku ro''' which adam mentioned, according to [[DeMorgan|DeMorgan]]).  '''me'iro''' isn't the same as this (I think [[User:xorxes|xorxes]] is responsible for starting using me'iro for this at the [[Existential Import|Existential Import]] page), and I agree with And:  there's no reason it shouldn't be allowed to equal 0.  However, it <u>does</u> claim that the cardinality of '''lo'i [[ro|ro]] broda''' is > 0, because it's false that 0 < 0.  The reason me'iro isn't up to the task of "some are not", is because if it is equal to zero, it is the same as "ro broda naku", which is not the same as the '''naku ro broda''' we're trying to say (it then means naku su'o broda, i.e. no broda). --mi'e [[.djorden.|.djorden.]]
* A sentence such as '''me'i broda cu brode''' could in fact be true if '''no broda cu brode''', but certainly not if there are no broda whatsoever. If there are 9 broda, and none of them are brode, then clearly '''me'i broda cu brode''' is true. If there are 0 broda (and thus none of them are brode) then '''me'i broda cu brode''' is false. Thus, it clearly has existential import, since it's false when there are no broda. If it didn't have existential import, then it would be vacuously true when there are no broda, which is what John and pc want. '''me'i broda cu brode''', just like '''su'o broda naku brode''', allows the possibility that '''ro broda naku brode''', given that there is at least one broda (for both sentences), but does not require it. I like '''me'i''', because it gives a nice symmetry; there is a one-word quantifier for each of the four basic quantifiers of categorical logic, but it doesn't make a big difference. [[User:xorxes|xorxes]] proposed '''me'iro''' in a previous round of this debate, and also proposed '''da'asu'o''' as a way to say ''not all'', i.e. ''all but at least one''. -- AdamRaizen
* Yes. Sorry. Your point had been obvious to me, but it evaporated from my confused mind when I wrote that bit. --[[User:And Rosta|And Rosta]]
* I don't follow his point.  '''me'iro broda''' can be 0 broda, but '''su'o broda naku''' cannot.  '''me'iro''' is <u>a</u> quantifier (which does import), but it is not the same as the 'O' quantifier.  I think '''da'asu'o''' works (because it can't be zero), but '''da'asu'o''' != '''me'iro''' - the [[Existential Import|Existential Import]] page, which also claims '''da'ame'iro''' == '''su'o''' (which, if '''me'iro''' is equal to 0 is actually the same as '''da'aro''', which is the same as no (i.e. '''naku su'o''') is wrong.  --mi'e [[.djorden.|.djorden.]]
** The problem with allowing '''me'iro broda''' to be equivalent to '''no broda''' is that then '''ro''' == '''no''', so you have '''me'ino''' reducing to '''no''' which is, at least, very strange, because it's n < 0 && n = 0, which most logics would be upset with.  '''My''' question is: is '''me'i broda''' == '''me'iro broda'''? -- RobinLeePowell
*** How do you go from '''me'iro''' == '''no''' to '''ro''' == '''no'''? Those two are clearly contradictory, you can't have both '''ro''' and '''me'iro''' being equal to '''no''' in the same case. Each of them can be '''no''' in some case, but not both at the same time. When '''ro''' == '''no''', '''me'iro''' gives FALSE, as it should be for an importing quantifier, while '''ro''' gives TRUE, as it should be for a non-importing quantifier. When '''me'iro''' == '''no''', then '''ro''' is of course '''za'uno'''. --[[User:xorxes|xorxes]]
**** Umm, you just agreed with me.  In the case in which there are no da zo'u da broda, you cannot allow people to ''also'' say me'iro broda, because of the contradiction that I stipulated and you repeated.  -- RobinLeePowell
*** Whether ''me'i broda'' is equivalent to ''me'iro broda'' is merely a matter of convention, just as the fact that ''su'o broda'' is equivalent to ''su'opa broda'' is a convention.
** ''su'o broda naku brode'' is certainly compatible with ''no broda cu brode''. "At least one doesn't" does not entail that "at least one does". --[[User:xorxes|xorxes]]
* A quantifier is said to have existential import if ''Q broda cu brode'' entails ''(su'o) da broda''. As Adam & xorxes have explained, ''me'i (ro) broda cu brode'' entails ''(su'o) da broda'' (even though it doesn't entail ''su'o broda cu brode''). OTOH, ''ro broda cu brode'' does not entail ''(su'o) da broda'' (and nor does it entail ''su'o broda cu brode''). --[[User:And Rosta|And Rosta]]
* For clarification: I haven't thought about whether ''me'i'' can serve as the 'O' quantifier. My remarks above were solely addressed to the question of whether ''me'i'' has existential import. --[[User:And Rosta|And Rosta]]
** I was dealing solely with whether it is the 'O' quantifier, and it certainly seems that calling it one of the 'principal quantifiers' above suggests that it was being claimed that it is (and it *certainly* is claimed that it is 'O' on the [[Existential Import|Existential Import]] page).  I agree that it obviously imports, because 0 < 0 is false (as I stated in my first comment), but it has to be deductively equivalent to ''su'o broda naku'' (and it's not) to be 'O'.  --mi'e [[.djorden.|.djorden.]]
*** In what instance is ''su'o broda naku brode'' true but ''me'iro broda cu borde'' false, or vice versa? Let us say that A = the denotation of broda (lo'i broda) and B = the denotation of brode (lo'i brode). ''su'o broda naku brode'' means that A \ B is not empty (i.e., there is some a in A which is not in B). If A and B are finite[[1|1]], then ''me'iro broda cu brode'' means that   @{#8745} B| < |A| (i.e., there are fewer things which are both A and B than there are things which are A). If A \ B is not empty, then A is not a subset of B, or in other words there is an a in A which is not in B. Thus, a is not in the intersection of A and B, and so the cardinality of A @{#8745} B is less than the cardinality of A (since the cardinality of A @{#8745} B can be at most the cardinality of A anyway). On the other hand, if |A @{#8745} B| < |A|, then just the opposite: there is some a in A which is not in B, and so A \ B contains a, and so is not empty. Therefore, A \ B @{#8800} @{#8709} <=> |A @{#8745} B| < |A|, and thus ''su'o broda naku brode'' is equivalent to ''me'iro broda cu brode''. -- [[Adam Raizen|Adam Raizen]]
**** You're right:  what was confusing me was the case (which And brought up above) in which A - B = A (i.e. the case where me'iro evaluates to 0: the intersection is empty so the complement is the same as A), however all su'o claims is that |A - B| >= 1, and thus it makes perfect sense.  Thanks for the excellent explanation.  --mi'e [[.djorden.|.djorden.]]

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[[1|1]] If A and B are infinite, then I am not sure how this would work. If you simply take it that ''ro'' is replaced with whatever is the actual cardinality of the set, then I suppose as statement like "Less than all natural numbers are multiples of 2" is false. However, if we consider ''ro'' to be just the cardinality of the underlying set, then ''ro rarna'u cu pilji li re'' is true, since there are @{#8501}-0 natural numbers which are multiples of 2, so perhaps considering ''ro'' to be whatever the cardinality of the underlying set is is a convenient shortcut, but not always strictly accurate. At any rate, for things in the world (which is finite) ''me'iro'' is equivalent to ''su'o ... naku''. -- AdamRaizen
* I think this problem goes away when ''ro'' is viewed as an iterative operator instead of as the cardinality of the set.  Then ''ro rarna'u cu pilji li re'' is false, because there are members of the set of rarna'u which for which the propositional function will evaluate false.  --mi'e [[.djorden.|.djorden.]]
** Yes, quite right. But then the problem is how to interpret ''me'iro'' when ''ro'' is an iterator. The obvious solution would be that the iterator would have to find some x in the set being iterated over for which the predicate is false, and that is just the same as ''su'o ... naku'' anyway. I guess I'll give And about a day as I think this over before I break the consensus on the other page. -- AdamRaizen
* Since 'inner ro' expresses a cardinality, it is simplest to say that ro always expresses a cardinality. However, you (Jordan) suggested construing all cardinals as iterators, so it is therefore not impossible to view ro as an iterator. It is important, though, to maintain that ro is a cardinal, since this was the only decisive argument in favour of its nonimportingness. --[[User:And Rosta|And Rosta]]
** I don't understand how to construe a cardinal as an iterator, so maybe you could explain that. While I agree that it would be good to construe ''ro'' the same in all circumstances, I don't see how it is possible, given the above. At any rate, the most straightforward algorithm for construing ''ro'' as an interator does not give it existential import, so the need for existential import is not a decisive factor. -- AdamRaizen