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Assume we are in Basic Logic introduced in here (fuzzy logic), and we have $\{\varphi\}\vdash \psi$, and $\{\neg \varphi\}\vdash \psi$. Can we conclude that we have $\vdash \psi$? Or do we have to provide a proof? Can I consider it as obvious? If we have to prove it, where should I start?

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    $\begingroup$ "Basic Logic" doesn't exist but you can derive that in any logic with a cut rule and the law of excluded middle. $\endgroup$ Commented Jun 24 at 12:09
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    $\begingroup$ @NaïmFavier I mean the basic logic introduced in chapter 2 of Metamathematics of Fuzzy Logic (link.springer.com/book/10.1007/978-94-011-5300-3) $\endgroup$
    – Doralisa
    Commented Jun 24 at 12:10
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    $\begingroup$ @NaïmFavier In nonclassical logic such as fuzzy logic we do not have $\vdash p \vee \neg p$, since we have $V(p \vee \neg p) = \max \{p, 1-p\}$. And if $V(p)=0.5$ we do not have $\vdash p \vee \neg p$. $\endgroup$
    – Doralisa
    Commented Jun 24 at 12:49
  • $\begingroup$ @NaïmFavier: your comment is a typical example where I am annoyed that MSE does not allow one to downvote comments. It is wrong. Please delete it. $\endgroup$
    – Rob Arthan
    Commented Jun 24 at 22:12
  • $\begingroup$ @RobArthan My comment was made before the question included a link or any reference to fuzzy logic, as if those words alone would unambiguously identify that logic from that book. Note that you can flag comments as "no longer needed" if it really itches you. $\endgroup$ Commented Jun 24 at 23:36

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Unfortunately, we can not deduce $\vdash \psi$ from $\{\varphi\}\vdash \psi$, and $\{\neg \varphi\}\vdash \psi\quad $ :(

I am really sad I wish we could have it in BL too. I could make the following counter-example:

Let $\varphi = p$, and $\psi = (p\&p)\veebar (\neg p\&\neg p)$, then we have

$$\{p\}\vdash (p\&p)\veebar (\neg p\&\neg p),\; and \quad \{\neg p\}\vdash (p\&p)\veebar (\neg p\&\neg p)$$ but $\nvdash (p\&p)\veebar (\neg p\&\neg p)$.

Note that $\&$ is the strong conjunction which its semantics is defined as $p\&q = \max\{0,p+q-1\}$. Also, $\veebar $ is the strong disjunction i. e. $p \veebar q = \min \{1, p+q\}$.

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