A set $\sum$ of formulas in propositional logic is complete if for each propositional formula $\phi$ either $\sum \vdash \phi$ or $\sum \vdash \neg \phi$. Clearly every inconsistent set of formulas is complete because of the following lemma
Lemma: Let $\sum$ be an inconsistent set, then for every propositional formula $\phi$ , $\sum \vdash \phi$
So the important thing is determining whether a consistent set of formulas is complete or not. I would like to know is there any method to find out whether a consistent set of formulas is complete or not?
As an example the following sets are complete ($\downarrow$ means NOR)
$\{p_1,p_1 \leftrightarrow p_2,p_2 \leftrightarrow p_3,p_3 \leftrightarrow p_4,… \}$
$\{p_1 \downarrow p_2,p_2 \downarrow p_3,p_3 \downarrow p_4,p_4 \downarrow p_5,…\}$
But this one is not
$\{\neg p_1 , p_1 \vee p_2,p_1 \vee p_2 \vee p_3,p_1 \vee p_2 \vee p_3 \vee p_4 , … \}$
UPDATE:
From what I tried, I got this
$$\{\neg p_1 , p_1 \vee p_2,p_1 \vee p_2 \vee p_3,p_1 \vee p_2 \vee p_3 \vee p_4 , … \} \nvdash p_1$$
Because when we write $$\{\neg p_1 , p_1 \vee p_2,p_1 \vee p_2 \vee p_3,p_1 \vee p_2 \vee p_3 \vee p_4 , … \} \vdash p_1$$
it means every model for the left side (satisfies every element of the set) must be a model for the right side, But there is no way to find a model to satisfy $$\{\neg p_1 , p_1 \vee p_2,p_1 \vee p_2 \vee p_3,p_1 \vee p_2 \vee p_3 \vee p_4 , … \} \vdash p_1$$
because every element of the left side must be $T$ so $\neg p_1$ must be $T$ so it means $p_1 = F$ it concludes that the right side is $F$, the same reasoning holds for $$\{\neg p_1 , p_1 \vee p_2,p_1 \vee p_2 \vee p_3,p_1 \vee p_2 \vee p_3 \vee p_4 , … \} \vdash \neg p_1$$ so I found the propositional formula $p_1$ such that $$\{\neg p_1 , p_1 \vee p_2,p_1 \vee p_2 \vee p_3,p_1 \vee p_2 \vee p_3 \vee p_4 , … \} \nvdash p_1$$ and $$\{\neg p_1 , p_1 \vee p_2,p_1 \vee p_2 \vee p_3,p_1 \vee p_2 \vee p_3 \vee p_4 , … \} \nvdash \neg p_1$$ so as a result $$\{\neg p_1 , p_1 \vee p_2,p_1 \vee p_2 \vee p_3,p_1 \vee p_2 \vee p_3 \vee p_4 , … \}$$ is not complete.But I was looking for a more algorithmic (even a semi-decidable one) to solve the problem.
Best Answer
So, to state what is probably obvious, a set of propositions is consistent if there is at least one assignment of truth values to literals that makes all the propositions true; and a set of propositions is complete if there is at most one such assignment. For the set of propositions to be both consistent and complete, then, there must be a unique satisfying assignment of truth values. To show it is not complete, then, you just need to exhibit two satisfying assignments. In your example, for instance, the assignments $$ \{F, T, T, T, T, \ldots \} $$ and $$ \{F, T, F, F, F, \ldots\} $$ both make all the propositions true.