Suppose that we define consistency in Łukasiewicz logic as follows:
- We say formula $\varphi$ is consistent if $\nvdash \neg \varphi$.
- We say a finite set $\Gamma= \{\psi_1, \dotsc, \psi_n\}$ is consistent if $(\psi_1\&\dotsb\&\psi_n)$ is consistent.
- We say an infinite set $\Gamma$ is consistent if all finite subsets of $\Gamma$ are consistent.
And also define maximality as follows:
- We say a set $\Phi$ is maximal if for all $\varphi\notin \Phi$, $\Phi\cup \{\varphi\}$ is inconsistent.
Then we are able to prove the following:
If $\Phi$ is a maximal and consistent set, then for all formulæ $\varphi$ and $\psi$:
- $\varphi\&\psi\in \Phi$ if and only if $\varphi\in\Phi$ and $\psi\in\Phi$,
- If $\varphi\in \Phi$ and $\varphi\rightarrow \psi\in \Phi$, then $\psi\in \Phi$,
- If $\vdash \varphi$, then $\varphi\in \Phi$,
- $\varphi\in \Phi$ or $\neg \varphi\in \Phi$.
Proof Sketches:
1:
Let $\varphi\&\psi\in \Phi$ and $\varphi\notin \Phi$. Thus by definition $\Phi\cup \{\varphi\}$ is not consistent. So there is a finite set $\Gamma\subset\Phi$ such that $\Gamma\cup \{\varphi\}$ is not consistent. Let $\Gamma=\{\psi_1,\dotsc,\psi_n\}$, hence by the inconsistency of $\Gamma\cup \{\varphi\}$ we have $\vdash \neg (\psi_1\&\dotsb\&\psi_n\&\varphi)$. Since $\Gamma\cup \{\psi,\varphi\}$ is not consistent, then we have $\vdash \neg (\psi_1\&\dotsb\&\psi_n\&\varphi\&\psi)$, and since $\Gamma\cup \{\varphi\&\psi\}\subset \Phi$, then we obtain a contradiction to the Aconsistency of $\Phi$.
For other direction assume $\varphi\in \Phi$, $\psi \in \Phi$ and $\varphi \,\&\, \psi \notin \Phi$. Then $\Phi \cup \{\varphi\,\&\, \psi\} $ is not consistent, and there is a finite set $\Gamma = \{\psi_1, \dotsc, \psi_n\}$, where $\Gamma\subset \Phi$ such that $\vdash\neg (\psi_1\,\&\, \dotsb\,\& \,\psi_n\,\&\, \varphi \,\&\,\psi)$. But this states that the finite subset $\Gamma \cup \{\varphi,\psi\} \subset \Phi$ is not consistent.
2:
Let $\varphi \in \Phi$ and $\varphi \rightarrow \psi \in \Phi$. If $\psi \notin \Phi$, then $\Phi \cup \{\psi\}$ is inconsistent and there exists a finite subset $\Gamma \subset \Phi$ such that $\Gamma \cup \{\psi\}$ is inconsistent. Let $\Gamma = \{\psi_1, \dotsc, \psi_n\}$, so we have:
\begin{equation}\tag{1}\label{eq_inconsistent} \vdash \neg (\psi_1\,\&\, \dotsb \,\&\, \psi_n \,\&\, \psi). \end{equation}
On the other hand, it can be proven that from $\vdash \chi\rightarrow \chi$ and $(\varphi\& (\varphi \rightarrow \psi)) \rightarrow \psi$ we have $\vdash(\chi \rightarrow \chi)\,\&\, ((\varphi \,\&\,(\varphi \rightarrow \psi)) \rightarrow \psi)$ and by considering an instance of $((\varphi_1 \rightarrow \psi_1) \& (\varphi_2 \rightarrow \psi_2)) \rightarrow ((\varphi_1 \&\varphi_2) → (\psi_1 \&\psi_2)))$ and applying (MP) we obtain $$\vdash\chi \,\&\, (\varphi \,\&\, (\varphi \rightarrow \psi)) \rightarrow (\chi \,\&\, \psi).$$ Then by replacing $\psi_1\,\&\, \dotsb \,\&\,\psi_n$ instead of $\chi$ we have: $$\vdash (\psi_1\,\&\,\dotsb\,\&\, \psi_n \,\&\, \varphi \,\&\,(\varphi \rightarrow \psi)) \rightarrow (\psi_1 \,\&\, \dotsb \,\&\, \psi_n \,\& \,\psi ).$$ Now, by using $(\varphi \rightarrow \psi)\rightarrow (\neg \psi \rightarrow \neg \varphi)$ scheme, which is provable in Łukasiewicz logic we have:
\begin{equation}\tag{2}\label{eq013} \vdash\neg (\psi_1 \,\&\, \cdots \,\&\, \psi_n \& \psi ) \rightarrow \neg (\psi_1\,\&\,\cdots\,\&\, \psi_n \,\&\, \varphi \,\&\,(\varphi \rightarrow \psi)). \end{equation}
By applying (MP) on \eqref{eq_inconsistent}, \eqref{eq013} we obtain $\vdash \neg (\psi_1\,\&\,\dotsb\,\&\, \psi_n \,\&\, \varphi \,\&\,(\varphi \rightarrow \psi)) $. But this means that $\Gamma \cup \{\varphi, \varphi\rightarrow \psi\}\subset \Phi$ is inconsistent, which is a contradiction to the consistency of $\Phi$.
3:
Let $\vdash \varphi$. By soundness in Łukasiewicz logic we have $\neg \varphi \equiv \bot$, which means that $\neg \varphi$ is semantically equivalent to $\bot$. If $\varphi \notin \Phi$, then by maximality definition $\Phi\cup \{\varphi\}$ is inconsistent. So there is a subset $\Gamma = \{\psi_1,\dotsc, \psi_n\} \subset \Phi$, such that $\Gamma\cup \{\varphi\}$ is not consistent. Therefore $$\begin{array}{llll} (1) & & \vdash\neg (\psi_1\,\&\, \dotsb \,\&\, \psi_n \,\&\, \varphi) & \text{inconsistency of } \Gamma \cup \{\varphi\}\\ (2) & & \vdash\neg \psi_1 \veebar\dotsb \veebar \neg\psi_n \veebar \neg \varphi & (1),\text{Ł}5 \\ (3) & & \vdash\neg (\neg \psi_1 \veebar\dotsb \veebar \neg\psi_n) \rightarrow \neg \varphi & (2), \text{Ł}7 \\ (4) & & \vdash\neg (\neg \psi_1 \veebar\dotsb \veebar \neg\psi_n) \rightarrow \perp & (3), \text{assumption and replacement} \\ (5) & & \vdash(\neg \neg \psi_1 \,\&\, \dotsb \,\&\, \neg \neg \psi_n) \rightarrow \perp & (4), \text{Ł}6 \\ (6) & & \vdash(\psi_1 \,\&\, \dotsb \,\&\, \psi_n) \rightarrow \perp & (5), \text{Ł}11\\ (7) & & \vdash\neg (\psi_1 \,\&\, \dotsb \,\&\,\psi_n) & (6),\text{Ł}15 \end{array}$$ Thus from $\vdash \neg (\psi_1 \,\&\, \dotsb \,\&\,\psi_n)$, we have $\Gamma$ is inconsistent. But this is a contradiction with the consistency of $\Phi$.
4:
Suppose that there is a formula $\varphi$ such that $\varphi\notin \Phi$ and $\neg \varphi \notin \Phi$. From the maximality and consistency of $\Phi$, neither $\Phi \cup \{\varphi\}$ nor $\Phi \cup \{\neg \varphi\}$ is consistent. Assume $\Gamma = \{\psi_1, \dotsc, \psi_m\}$ is a finite subset of $\Phi$ for which $\Gamma \cup \{\varphi\}$ is inconsistent, then the following deduction is valid: $$\begin{array}{llll} (1)& &\vdash \neg (\psi_1 \,\&\, \dotsc \,\&\, \psi_m \,\&\,\varphi) & \text{inconsistency of } \Gamma\cup \{\varphi\}\\ (2)& & \vdash \neg \psi_1 \veebar \dotsb \veebar \neg \psi_m \veebar \neg \varphi & (1), \text{Ł} 5\\ (3)& &\vdash \neg (\neg \psi_1 \veebar \dotsb \veebar \neg \psi_m ) \rightarrow \neg \varphi & (2), \text{Ł}7 \\ (4)& &\vdash (\neg\neg \psi_1 \,\&\, \cdots \,\&\,\neg\neg\psi_m ) \rightarrow \neg \varphi & (3), \text{Ł}6 \\ (5)& & \vdash (\psi_1 \,\&\, \cdots \,\&\,\psi_m ) \rightarrow \neg \varphi & (4), \text{Ł}11. \end{array}$$ Thus from 3 we have $(\psi_1 \,\&\, \dotsb \,\&\, \psi_m )\rightarrow \neg \varphi \in \Phi$, since $\Gamma=\{\psi_1, \dotsc, \psi_m\}\subset \Phi$ then we have $(\psi _1 \,\&\, \dotsb \,\&\, \psi_m) \in \Phi$, by 1. So from 2 we have $\neg \varphi \in \Phi$, which is a contradiction with the assumption $\neg \varphi \notin \Phi$.
My question is about the strange set $\{\neg(x\& x), \neg(\neg x\&\neg x)\}$. It is easy to see that this set is consistent, and so it can be extended to a maximal and consistent set $\Theta$. By maximality of $\Theta$ and the above statements (Number 4), we have $(x\& x)\notin \Theta$ and $(\neg x \& \neg x) \notin \Theta$. So by number 1, $x\notin \Theta$ and also $\neg x\notin \Theta$. But this can not be possible. Why does this happen? What is the problem with this consistency definition and the maximality? Or am I missing something important?
Edit: By $\&$ we mean strong conjuction. That its semantics informally is $V(x \& y) = \max\{0, x + y -1\}$.
Also, the Łukasiewicz theorems used in the proofs are as follows: $$\begin{array}{llcll} (\text{Ł}5) & \neg (\varphi \,\&\, \psi) \leftrightarrow (\neg \varphi \veebar \neg \psi) & \quad & (\text{Ł}6)& \neg (\varphi \veebar \psi) \leftrightarrow (\neg \varphi \, \& \, \neg \psi) \\ (\text{Ł}7)& (\varphi \veebar \psi) \leftrightarrow (\neg \varphi \rightarrow \psi) & \quad&(\text{Ł}11) & \neg \neg \varphi \leftrightarrow \varphi\\ (\text{Ł}15) &(\varphi \rightarrow \perp) \leftrightarrow \neg \varphi & & & \end{array}$$