Is ⟨me'i⟩ importing?: Difference between revisions
No edit summary 
m (Gleki moved page is me'i Importing to Is ⟨me'i⟩ importing?) 
(No difference)

Latest revision as of 12:40, 27 June 2019
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. 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). me'iro isn't the same as this (I think xorxes is responsible for starting using me'iro for this at the Existential Import page), and I agree with And: there's no reason it shouldn't be allowed to equal 0. However, it does claim that the cardinality of lo'i 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.
 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 oneword quantifier for each of the four basic quantifiers of categorical logic, but it doesn't make a big difference. 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. And Rosta
 I don't follow his point. me'iro broda can be 0 broda, but su'o broda naku cannot. me'iro is a 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 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.
 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 nonimporting quantifier. When me'iro == no, then ro is of course za'uno. 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.
 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 nonimporting quantifier. When me'iro == no, then ro is of course za'uno. xorxes
 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". xorxes
 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
 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). 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. 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 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.
 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 finite1, 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
 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.
 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 finite1, 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
 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 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.
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.
 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. 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