Existential import

We consider the following logical forms:

A All S is P

E No S is P

I Some S is P

O Not all S is P

We can divide each of these in two cases, according to whether or not they have existential import, i.e. whether or not they entail that there is some S. We indicate import with + and no import with -, so we have:

A- All S, if there is any S, is P

A+ All S, and there is some S, is P

E- No S, if there is any S, is P

E+ No S, and there is some S, is P

I- Some S, if there is any S, is P

I+ Some S, and there is some S, is P

O- Not all S, if there is any S, is P

O+ Not all S, and there is some S, is P

The non-controversial extended Lojban forms for these are:

A- roda zo'u ganai da broda gi da brode

E- noda zo'u ge da broda gi da brode

I+ su'oda zo'u ge da broda gi da brode

O+ me'ida zo'u ganai da broda gi da brode

A+ ge da broda gi rode zo'u ganai de broda gi de brode

E+ ge da broda gi node zo'u ge de broda gi de brode

I- ganai da broda gi su'ode zo'u ge de broda gi de brode

O- ganai da broda gi me'ide zo'u ganai de broda gi de brode

For the first four, the import condition is not needed, because it is already contained in the second part.

The controversial (as to their import) short Lojban forms are these:

A ro broda cu brode

E no broda cu brode

I su'o broda cu brode

O me'i broda cu brode

We should note the relationships that exist between forms:

A- = E- naku = naku I+ naku = naku O+

A+ = E+ naku = naku I- naku = naku O-

These relationships tell us how to change the quantifier when going through a negation boundary. We see that there is a natural grouping of (A-E-I+O+) and (A+E+I-O-). The fisrt group is the one that does not require an explicit import condition in the extended forms.

In principle, any assignment could be made to the short forms. For xorxes e, the correct assignment is (A-E-I+O+), as it is the simplest and most basic. Others prefer (A+E+I+O+) or (A+E-I+O-). These introduce complications when you try to relate one form to another through negations. So this is how xorxes understand them, but keep in mind that this assignment is not accepted by everyone:

A- ro broda cu brode

E- no broda cu brode

I+ su'o broda cu brode

O+ me'iro broda cu brode

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Complementary relationships between the quantifiers:

da'a ro = no

da'a no = ro

da'a su'o = me'i

da'a me'i = su'o

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All four quantifiers can also be written in terms of pa or da'apa:

ro = za'u da'apa

no = me'i pa

su'o = su'o pa

me'i = su'e da'apa

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Two other not very useful but interesting forms are the tautology and the contradiction, which I like to call U and Y just to use the remaining Lojban vowels:

U su'ono broda cu brode (At least zero S are P - tautology)

Y me'ino broda cu brode (Less than zero S are P - contradiction)

The natural import for those is U- and Y+, as can be seen from the extended forms:

U- su'onoda zo'u ganai da broda gi da brode

U- su'onoda zo'u ge da broda gi da brode

Y+ me'inoda zo'u ge da broda gi da brode

Y+ me'inoda zo'u ganai da broda gi da brode

U+ ge da broda gi su'onode zo'u ganai de broda gi de brode

U+ ge da broda gi su'onode zo'u ge de broda gi de brode

Y- ganai da broda gi me'inode zo'u ge de broda gi de brode

Y- ganai da broda gi me'inode zo'u ganai de broda gi de brode

And we also have the relationships:

U- = U- naku = naku Y+ = naku Y+ naku

U+ = U+ naku = naku Y- = naku Y- naku

These quantifiers can also be written in terms of ro instead of no:

su'ono = su'ero

me'ino = za'uro

And they are complementary:

da'a su'ono = me'ino

da'a me'ino = su'ono

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Another interesting quantifier quartet is given by rau (enough) du'e (too many) and mo'a (too few). We don't have a cmavo for the fourth one, "not too many". Perhaps we can use {me'idu'e} for it. These follow their own DeMorgan's transformations:

du'eda = mo'ada naku = naku rauda naku = naku me'idu'eda

mo'ada = du'eda naku = naku me'idu'eda naku = naku rauda

rauda = me'idu'eda naku = naku du'eda naku = naku mo'ada

me'idu'eda = rauda naku = naku mo'ada naku = naku du'eda

Then the complementaries would be:

da'adu'e = mo'a

da'amo'a = du'e

da'arau = me'idu'e

da'ame'idu'e = rau

xorxes