## Monday, February 29, 2016

### What can we learn from the Contingent Liar?

• The last bulleted item in this post is not true.
Call this token bulleted linguistic item p. Then p is, in fact, the last--and the only--item prefixed with a bullet point in this text. If it's not true, then it seems it's true. But if it's true, then it seems it's not. Oops! That's a contradiction in classical logic. But, famously, sentences like this are only contingently paradoxical. If I were to add a bullet point followed by "2+2=4" at the end of the post, then p would be unparadoxically false, while if I were to end the post with a bullet point followed by a piece of nonsense or a falsehood, then p would be unparadoxically true (nonsense is not true).

Here is an assumption that I think is implicit in the above derivation:

1. The item p is true if and only if the last bulleted item in this post is not true.
The argument needs some bridge like this between the truth of the linguistic item p and the last bulleted item not being true. Once we have (1), then the argument is quick, using only uncontroversial premises. If item p is true, then the last bulleted item is not true by (1). But empirically the last bulleted item is p. So if p is true, then it's not true. But if it's not true, then by (1) it's not the case that the last bulleted item is not true. Since empirically the last bulleted item is p, it follows that it's not the case that p is not true, i.e., that p is true. So p is true if and only if it's not true, a contradiction in classical logic.

Since we should not deny classical logic or obvious empirical truths, it follows that (1) is false. Now, if p expresses a proposition, then it's got to express a proposition that makes (1) be true--that's both intuitively obvious and a consequence of the Tarski T-schema. (Doesn't (1) follow from the T-schema absent the expression assumption? It had better not. If "s" is meaningless, then the instance "'s' is true if and only if s" does not express a proposition, too, and hence is not true. So the T-schema had better apply only to meaningful items.) So p doesn't express a proposition. But that's a contingent fact, since in another possible world I screw up and end this post with a bulleted "2+2=5" thereby making p both meaningful and true.

So, whether a linguistic item expresses a proposition is in general a contingent matter. We already should have known this in the case of linguistic items using names, indexicals and demonstratives, and indeed p contains the demonstrative "this". But nothing hangs on p containing the demonstrative "this"--one could just replace it with some complex definite description--so I will ignore this demonstrative. If we think that whether a linguistic item expresses a proposition determines whether it's a meaningful sentence, then it follows that whether a linguistic item is a meaningful sentence is contingent, even in the absence of names, indexicals and demonstratives.

Further, not only is it a contingent matter whether a linguistic item expresses a proposition, but whether it does so can vary from token to token, again in the absence of names, indexicals and demonstratives. After all, I just gave a conclusive argument p does not express a proposition, and hence that the last bulleted item in this post does not express a proposition, and thus is not true:

1. The last bulleted item in this post is not true.
Item token (2) is true (note that numerals aren't bullets) and hence expresses a proposition. But s, which is a token of exactly the same type, does not. And that's not due to names, indexicals or demonstratives.

These conclusions are interesting independently of the paradox. But somehow it feels wrong to use the paradox to reach them. Is it?