Tarski's T-schema says that for any sentence "s":
- "s" is true if and only if s.
It's odd that I never noticed this before.
Thursday, June 9, 2011
T-schema and bivalence
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Tarski's T-schema says that for any sentence "s":
It's odd that I never noticed this before.
4 comments:
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? ;)
Some posts are just notes to self. :-)
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.
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?
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