Comments on: How many is two?

By: Thad Coons

Thad Coons — Fri, 27 May 2005 17:15:46 +0000

The failure of many rules of inference does appear to be of the strongest arguments against Lukasiewicz 3-valued logic. But apparently no one has noticed that it’s possible to develop his logic in a way that does establish these rules of inference given certain reasonable conditions. My results have some features that are much like intuitionism. Since I’m an amateur with radically unorthodox ideas and not an expert on intuitionistic logic, I’m interested in comparing them. Please bear with me.

By: Andrej Bauer

Andrej Bauer — Fri, 27 May 2005 15:14:30 +0000

You are correct. Bivalence is not valid in intuitionistic logic, whereas noncontradiction is.

There is a lot of confusion about what it means to have two-valued logic (which is why I wrote this post). Lukasiewicz’s three-valued logic is not a model of intuitionistic logic, because some rules of inference are not valid in it. I was not talking about just any arbitrary thing that anyone might call logic, but rather specifically about intuitionistic logic. But that does not matter since we can find models of intuitionistic logic which seemingly have “three values”, for example the topos of presheaves on the category consistig of two objects and an arrow between them. In fact, given any Heyting algebra `H` with as many points as you like, you can find a model of intuitionistic logic, namely the `H`-valued sets, such that `Omega` is the set `H` (with a suitably defined `H`-valued equality predicate). But now it is important not to confuse the question “How many points does `H` have?” with the question “Inside the model, how many points does `Omega` have?”. Inside the model you can prove the validity of the statement `not exists p in Omega . (p != TT and p != _|_)`, but in the metatheory you can easily find a Heyting algebra `H` with more than two points. Please do not confuse a theory with meta-theory.

By: Thad Coons

Thad Coons — Thu, 26 May 2005 19:27:58 +0000

I have understood the law of the excluded middle as having two classically equivalent formulations: bivalence (p v ~P) and noncontradition ~(p & ~p). It appears to me that you are rejecting bivalence but accepting noncontradiction.
I’m astonished to learn that I cannot have three logical values, since Lukasieiwicz invented his 3-valued logic in 1920, and I’ve been working with it for some 20 years myself. Could you reference the proof?
If you’re talking about the proof in coq, I believe this is likely to include some implicit assumptions that I might not accept.

By: Andrej Bauer

Andrej Bauer — Sun, 22 May 2005 11:38:40 +0000

I am not going to reply to every comment here, but please do not take that as a sign of my agreeing or disagreeing with you. This blog is not intended to be a tutorial in logic. I shall also write more specialist topics.

Reply to Thad: you cannot have three logical values. There is a proof of this. And if you read my post again, you will discover that I do reject the Law of Excluded Middle, and I do tell you how you can regain it by considering only the decidable propositions.

Reply to roconnor: you are confusing syntax and semantics. I was very careful to say that a truth value is represented by a sentence, but did not claim that every truth value can be represented by a sentence, which is an assumption implicit in your post. The Lindenbaum algebras you mention are made up of only those truth values that are definable, i.e., represented by some expression in the logic. Also, truth is not the same thing as provability. One last remark: what makes you think that in constructive mathematics every set has a cardinality?

By: roconnor

roconnor — Sun, 22 May 2005 07:55:00 +0000

I disagree with the premis of this post. It clearly the case that the cardinality of the set of equivalence classes of logically equivalent sentences is greater than two. Even after you mod out by the axioms of ZFC, you still have sentences such as CH and Con(ZFC) which sit strictly between ⊤ and ⊥. In fact this set forms a rich boolean algebra in classical logic. The proof that &not(&notA ∧ ¬¬A) is straight-forward. Assume that both ¬A and ¬¬A are provable. Then applying the first to the second we get a proof of ⊥ as required. The proof in Coq is:

fun (A : Prop) (H : (A -> False) /\ ((A -> False) -> False)) =>
match H return False with
| conj B C => C B
end
     : forall A : Prop, ~ (~ A /\ ~ ~ A)