Is there any three valued logic with this truth tables?

Question

T AND T = T
T AND U = T
T AND F = F
F AND T = F
F AND U = F
F AND F = F
U AND T = T 
U AND U = U 
U AND F = F

T OR T = T
T OR U = T
T OR F = T
F OR T = T
F OR U = F
F OR F = F
U OR T = T 
U OR U = U 
U OR F = F

NOT T = F
NOT U = U
NOT F = T

T => T = T
F => F = U
T => F = F
F => T = U

Add the remaining for implication:

U => U = U
T => U = U
F => U = U
U => T = U
U => F = U

I'm not very sure about the last two. But if I miss them before I guess it is because they don't add "knowledge".

The idea of U is to meant NOTHING or VOID a thing that doesn't provide any knowledge and from where we cannot obtain any information.

What is the worst paradox this logic could generate?

Edit just to add another concrete interpretation I have when deriving the truth tables. Imaging an expression that runs forever. This espression is VOID because I'm not able to evaluate its truthness.

But probably this is total garbage. Maybe I'm just mixing inconsistents intuitions in my head. If this thing doesn't have a name I have little expectations...

Edit: After some time, my current me thinks the ultimate reason of the question is I don't like LEM standard intepretation. To divide the world in true and false and nothing else. Giving formal system limited powers to figure out thruth I find LEM inapropiate as an axiom.

Is there a name for this logic? A Wikipedia link etc... On three valued logic I saw Kleene logic but I dislike theirs truth tables... — Eduard, Jul 05 '19 at 20:47
Is your objection to Kleene the fact that False OR Undef. = Undef.? — scoopfaze, Jul 05 '19 at 20:50
You haven't fully explained how implication is handled - what if a "U" is involved somewhere? — Noah Schweber, Jul 05 '19 at 20:52
Incidentally I don't really see how to justify "U or U = U" and "U or F = F" simultaneously. If the idea is that U is something we don't yet know, then "U or F" should equal U. But if the idea is that we remove clauses involving U, then "U or U" becomes the empty disjunction which arguably should equal F (since there is no clause of the empty disjunction which is assigned "true") - and similarly, under this idea we should have "U and U" = T (every clause in the empty conjunction is true). — Noah Schweber, Jul 05 '19 at 20:54
@Eduard Why don't you like T and U = U? (I think you should say a bit more about how precisely you envision U behaving, intuitively.) — Noah Schweber, Jul 05 '19 at 20:55
U means VOID. VOID doesn't provide you any thruthness of falseness to you who are evaluating the expression. U or F means F because U has NOTHING for you. — Eduard, Jul 05 '19 at 20:57
@Eduard But under that interpretation we should have "U or U" be F, and "U and U" be T, for the reasons stated above. And again, you haven't finished the table for implication ... — Noah Schweber, Jul 05 '19 at 21:00
Under my totally subjective interpretation when I try to understand an expression that says "VOID or VOID" for me is "VOID" — Eduard, Jul 05 '19 at 21:16
I guess I just don't understand your intuition for U. Why should "U implies U" and "U implies T" evaluate to U? I certainly don't see any justification for the latter, unless we're thinking of U as something like "possibly true, possibly false," at which point we (again) throw rules like "T and U is T" into question. — Noah Schweber, Jul 05 '19 at 21:16
@NoahSchweber: The AND/OR/NOT rules look like they're motivated by an intuition where $U$ says "imagine this part of the formula was not here at all". — hmakholm left over Monica, Jul 05 '19 at 21:52
@HenningMakholm "The AND/OR/NOT rules look like they're motivated by an intuition where says "imagine this part of the formula was not here at all"." Yes, but then I don't see how to justify for example "U implies T is U." (Your last sentence, though, overstates the situation in my opinion: a classically-correct $\forall x(p(x)\rightarrow q(x))$ would here be interpreted as an infinite conjunction of Ts and Us, which - as long as there's a single T - would evaluate to T here. Of course the issue is what happens if no $x$ satisfies $p$ ...) — Noah Schweber, Jul 05 '19 at 21:52
@NoahSchweber: Neither did I at first, but actually it does seem to make sense if we consider that $\to$ makes most sense in the context $\forall x(p(x)\to q(x))$. In that case $p(x)\to q(x)$ becomes $U$ when $x$ doesn't satisfy $p$, and therefore presumably drops out of the interpretation of $\forall$, so we actually get the right truth value at the end -- except that vacuous truths are no longer true but "void". [Note: some crosstalk happend here because I only realized this after posting my first comment, and hastily removed a part that said the opposite of this]. — hmakholm left over Monica, Jul 05 '19 at 21:54
@HenningMakholm It still does cause a serious problem, though. Consider e.g. "For all $x$, if $x$ is a counterexample to Fermat's last theorem then the Frey curve corresponding to $x$ is non-modular" - that would evaluate to "U" here, even though it really, really shouldn't. It is "vacuous at the bottom" but for highly nontrivial reasons (incidentally, this subjective-but-important distinction between vacuous truths is one reason I support the material implication). — Noah Schweber, Jul 05 '19 at 21:57
@NoahSchweber: I'm not saying this is a good idea, only that I can see some kind of motivation for attempting it -- in the hope that it would do what we need and still avoid the conceptual problems many learners have with material implication. — hmakholm left over Monica, Jul 05 '19 at 22:00
@Eduard In that case - again - how do you justify "$U\implies T$ is $U$"? — Noah Schweber, Jul 05 '19 at 22:00
@Noah: agreed that the last two implications U implies T is U and U implies F is U are not crystal clear. At some time I write it down as T and F respectively, but then I change my mind. Again the original question was if this "logic" has a name or a wiki link from which I can learn more... — Eduard, Jul 05 '19 at 22:00
how do you justify " U⟹T is U" "? Again not sure at all, but what I think is because I cannot evaluate the first expression (it is still running forever) I obtain no knowledge. So the implication is VOID. — Eduard, Jul 05 '19 at 22:04
Not sure if it'll bug you ,but you've lost distributivity: T AND (F OR U) $\neq$ (T AND F) OR (T AND U) — Ali, Jul 05 '19 at 22:15
I think I see AND, OR and NOT in a diferent level than IMPLIES when evaluating EXPRESSIONS looking for its "truesness". — Eduard, Jul 06 '19 at 07:39
@Ali: Yes. A little bit. I need to think a little bit about that. I think an usual proving strategy could be restricting to EXPRESSIONs that make sense (ie true or false) doing standard logic and study the U case separately. — Eduard, Jul 06 '19 at 07:49

hamam_Abdallah · Answer 1 · 2019-07-05T22:51:06.667

-1

Logically speaking, we should have an equivalence between $$P \implies Q $$ and $$not Q \implies not P$$ but in your three valued one,

$$T \implies T ( =T)$$ is not equivalent to $$F \implies F (=U)$$

edited Jul 05 '19 at 22:51

answered Jul 05 '19 at 21:00

hamam_Abdallah

62,951

This question is very imprecise - but whatever the question means, I don't think this is an answer to it! – Alex Kruckman Jul 05 '19 at 21:40
@AlexKruckman I agree that the last line is a nonsequitur, but the part about the contrapositive does address the OP's final question "What is the worst paradox this logic could generate?". – Noah Schweber Jul 05 '19 at 22:01
If you change the semantics, there's no reason you "should have" any particular logical equivalence. You could reframe your answer as "one equivalence this semantics doesn't validate is ..." but even then, so? Why is this particular equivalence important? This equivalence doesn't even hold for constructive logic. There are many other classical and constructive equivalences that aren't validated by this semantics, e.g. $\top\land\varphi\equiv\varphi$. – Derek Elkins left SE Jul 07 '19 at 03:17

Is there any three valued logic with this truth tables?

1 Answers1