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EDIT:

We have three kinds of famouse fuzzy logic name: Godel, Luaksiewicz and product logic (see 1). We can define infinite valuation semantics for each of them in $[0,1]$, i. e. we can define a valuation function $\pi$ that assign $x\in[0,1]$ to an atomic proposition $p$ ($V(p)=x$). It is conventional to use $\vDash \varphi$ to define a valid formula $\varphi$, which means that for all valuation function $\pi$ we have $V(\varphi)=1$ (The maximum possible value we can have that is 1 which informally equal to TRUE in classical logic).

So, we can for example define a valuation function $\pi:\mathcal{P}\rightarrow [0,1]$, where $\mathcal{P}$ is set of atomic proposition, that $\pi$ assigns value $0.25$ to an atomic proposition $p$ and denote it by $V(p)=0.25$.

Strong conjuction is defined as $V(\varphi \,\&\, \psi) = \max\{0, V(\varphi)+V(\psi)-1\}$, and for definition of other connective you may see 1.

The formula such as $p \,\&\, \neg p$ is fixed value formula in infinit-valued (fuzzy) logic that for ALL possible $\pi$ we have $V(p \,\&\, \neg p)=0$.

My question is can we define a formula $\psi$ (i. e. using atomic proposition $p$) such that for all possible valuation function $p$ we have $V(\psi)=0.5$ (or any other fixed value in $(0,1)$)?

-------- What I have asked previously before edit:

I wonder can we construct a formula that always evaluate as 0.5 when we have infinite possible values belogns to [0,1]?

Actually, I guess there is no such formula, but I am not able to prove it. I have tried induction but I got stuck in the case of strong conjuction, that is when we have $\max\{0, x+y-1\}$, how can we be sure that combination of values of $x+y$ is not equal to 1.5 :(

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  • $\begingroup$ I'm just as confused as you. Can you please reframe your question in a better manner? $\endgroup$
    – Gwen
    Commented May 18 at 19:46
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    $\begingroup$ Each of the following formulas evaluates to $0.5$ for every $x$ in the interval $[0,1]$: $\;$ $0.5 + 0 \cdot x\;$ and $\;\max\{0.5, \, 0 \cdot x\}\;$ and $\;\min\{0.5, \, x+2\}\;$ and $\;{(0.5)}^{0 \cdot x + 1}.\;$ Many other such examples are easily found. $\endgroup$ Commented May 18 at 19:58
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    $\begingroup$ The close voters should bear in mind the possibility that they don't understand the clues in the question: particularly those in the tags. The brusque "I don't get it, so let's close it down" position is not defensible. Ask for the OP to clarify instead. $\endgroup$
    – Rob Arthan
    Commented May 18 at 21:09
  • $\begingroup$ @Rob Arthan: For what it's worth, I didn't vote to close, but I do think some additional background exposition is needed, even if only to cite some papers, books, web pages, wikipedia articles, or whatever to give an idea what the context is. Also, connoisseurs of a niche topic should be aware (surely from past experience in discussing the topic with others, if nothing else) that some clarification is likely needed to avoid an unintended interpretation, which I suspect the OP's first sentence is guilty of. $\endgroup$ Commented May 19 at 4:31
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    $\begingroup$ @Gwen Well my question get closed before 24 hours and even a chance I see the comments! I added mote information but I thought anyone who works fuzzy logic or have seen infinite valued logics or Godel logic and ... would understand my question. $\endgroup$
    – Doralisa
    Commented May 19 at 18:16

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It sounds like you are working in one of the standard frameworks for a formulation of fuzzy logic where the truth values are real numbers in the interval $[0, 1]$. Induction won't help you prove that there is no formula that always evaluates to $0.5$. Instead, think about what such a formula could evaluate to in classical logic: it has to evaluate to a classical truth value: $0$ or $1$ and not $0.5$.

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  • $\begingroup$ Yes. As I edited my question Godel, Lukasiewicz , ... are kinds of fuzzy logic that you can asisgn values [0,1] to formula. Umm, I did not understand your hint in the last sentence well, but thanks anyway. I will think about it. $\endgroup$
    – Doralisa
    Commented May 19 at 18:27
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    $\begingroup$ Here's some more detail: let's call an assignment $V$ classical if $V(p)\in \{0, 1\}$ for every atomic proposition $p$, Then, as you can show by induction on the structure of $\psi$, $V(\psi) \in \{0, 1\}$ for every formula $\psi$ and every classical assignment $V$. $\endgroup$
    – Rob Arthan
    Commented May 27 at 19:25

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