Thursday, June 9, 2011

T-schema and bivalence

Tarski's T-schema says that for any sentence "s":

  1. "s" is true if and only if s.
Suppose that "s" is neither true nor false. Then the left hand side of (1) is false, but the right hand side is neither true nor false. It seems to me that a reasonable multivalent logic will not allow an "if and only if" sentence to be true when one side of it is false and the other side is not false. So, it seems that the T-schema requires bivalence.

It's odd that I never noticed this before.

4 comments:

Icaras Triakis Harimau said...

Im just saying, you are the owner of my favorite blog on the entire internet, and i can barely grasp any of this stuff. grasp, wasnt that what your last post was about? ;)

Alexander R Pruss said...

Some posts are just notes to self. :-)

45e34d74-9e85-11e0-a138-000bcdcb8a73 said...

I am curious about your first inference. Certainly, if "s" is neither true nor false, then "'s' is true" is not true. But what reason do you have for concluding from this that it is false as opposed to being neither true nor false? In my mind, what holds for "s" holds for "'s' is true," no? If so, and given that "s" is neither true nor false, then "'s' is true" is neither true nor false as well.

Alexander R Pruss said...

Yeah, Jon Kvanvig suggested the same thing to me.

I think this leads to this iteration:

's' is ntnf (neither true nor false)
"'s' is ntnf" is ntnf
...

So, ntnf iterates.

One downside of this is that if you think truth, or evidence of truth, is needed for assertion, you can't say of any sentence that it's an example of a sentence that is ntnf. But without an example of a sentence that is ntnf, why think there are any?