Hello together,
I have a rather basic issue on propositional logic: first, consider an arbitrary set of formulas $T$ that is consistent and complete, i.e., for every propositional formula $\varphi$, either there holds $T\vdash\varphi$ or $T\vdash\neg\varphi$.
My Question is: Is there a minimal consistent and complete subset $T^\prime\subseteq T$, i.e., $T^\prime$ is still complete, but no proper subset $T^{\prime\prime}$ of $T^\prime$ is?
I know some examples where that is the case, namely e.g. the set of propositional variables $T_1=\{A_0,A_1,\dots\}$. One can also vary this example by considering any subset $T_2$ of the literals (i.e., variables or negations of them) such that for any $i\geq0$, either $A_i\in T$ or $\neg A_i\in T$. In these cases, both $T_1$ and $T_2$ are already minimal in the above sense.
However, my attempts to prove the question above failed so far. My first approach was it to successively eliminate formulas $\varphi$ that can be proven from the rest, i.e. such that $T\setminus\{\varphi\}\vdash\varphi$. More detailed: let $T=\{\varphi_0,\varphi_1,\dots\}$. Then we define a sequence $(T_i)_{i\geq0}$ of subsets of formulas of $T$ as follows:
$T_0:=T$
If $T_i$ is defined, then search if there is an index $j\geq0$ such that $T_i\setminus\{\varphi_j\}\vdash\varphi_j$. If yes, pick the least such $j$ and define $T_{i+1}:=T_i\setminus\{\varphi_j\}$. Otherwise, set $T_{i+1}=T_i$
Now if the construction stops at some point, i.e. if there is an $i\geq0$ such that $T_i=T_{i+1}$, then for least such $i$, $T_i$ is minimal by construction. However, if the construction runs forever, then we cannot argue that the intersection $\bigcap_{i\geq0}T_i$ is still complete because the notion of proof is finite.
I have made now several other attempts, including infinite proofs and axiom of choice, but still nothing seemed to helped. Maybe I have overseen something. Still, I think that the answer to the question is Yes. Does somebody have an idea (or have a counterexample) for this question.
Yours sincerely,
Martin