How to find out if a set of Propositional formulas is complete? 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.
 A: 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.
A: I suspect the terminology (complete set of formulas) is in the context of the completeness theorem for propositional calculus. In this context it means this:

Let Σ be a set of formulas in a language L. Σ is complete for L if it is consistent and for each formula [literal] φ in L, exactly one of φ and ¬φ belongs to Σ.

So, unless you spell out what the language (''L'') is, you cannot say if a set of propositional formulas is complete or not.
Basically the same def appears in another book.
Also, your example $\{ p_1, p_2 , p_3,\neg (p_1 \vee p_2 \vee p_3)\}$ is not consistent (so not complete according to the above) because it implies both $p_1$ and $\neg p_1$.
