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| In a message dated 10/20/2002 2:09:54 PM Central Daylight Time, jjllambias@hidden.email writes: << >Whatever it is it is to explain, it? >> Axiom Schema 4: << [broda lo'e brode] = [kairbroda tu'o ka ce'u broda] >> << , the property of being identical >with lo'e broda is the same as the property of being identical with lo >broda >(some broda, that is). But, if the proeprties are the same, then what has >those properties must be the same (whether we take properties in intension >or >extension) Yes. >and, since each of these critters obviously has the property of >being identical with what it is, What do you mean by "each of these critters"? The two expressions? Neither of the two expressions has the property in question. >> Well, everything has the property of being what it is, but the critters in question are lo broda and lo'e broda -- the things, not the expressions. The first of these has the property tu'o ka ce'u du lo broda, and the second tu'o ka ce'u lo'e broda. But since these properties are the same, the first, lo broda, has the property tu'o ka ce'u du lo'e broda. Therefore, lo broda du lo'e broda. << The two expressions are not identical. The underlying set is the same for both expressions. >> See immediately above. By the "underlying set" I suppose you mean lo'i broda. This is obvious, but beside the point. << It leads to a false result only if you apply {lo'e broda} = {lo broda}, which is false and not entailed by my definitions. >> Alas, it is entailed by your definitions and is false. And, if you don't like the proof above, consider the next one: that {brode lo broda} is materially equivalent to {brode lo'e broda}, so that, even if {lo broda} were not identical to {lo'e broda}, sentences involving {lo'e} -- in your sense -- would be redundant. Or the last proof, that your system makes {da poi broda zo'u mi sisku le ka ce'u du da} is materially equvalent to {mi sisku le ka da poi broda zo'u ce'u du da}, which is demonstrably false in real cases. |