Why these words are not formulas? First of all I know that exist a question similar to this, but I don't understand the answer.
Well my question is from the book "Mathematical Logic: Part I" and Chapter 1.1.1 where they talk about propositional formulas.
The question is: They give me the alphabet $A$ and the set of words $W(A)$ and a subset $P \subset W(A)$ with $a,b, c \in P$. The book say that $a, \big{(} a \implies (b \iff c)\big{)}$... are formulas but after this the bool says that $a \wedge b, \neg (a)$... are not formulas but they are words.
But reading the book I assume that $F_0 = P$ and for each $n$, $F_{n+1} = F_n \bigcup \{ \neg F: F \in F_n \} \bigcup \{ (F \alpha G): F,G \in F_n, \alpha \in \{ \wedge, \vee ,\implies, \iff \}   \}$
So if $F_0 = P = \{ a,b,c \}$ then $F_1 = F_0 \bigcup \{ \neg a, \neg b, \neg c\} \bigcup \{a  \wedge b,...\}$ so $a \wedge b \in F_1$ and with the definition of the set of propositional formulas constructed from $P$ where $\mathbb{F} = \bigcup_{n \geq 0} F_n$ so $a \wedge b \in \mathbb{F}$ and it will be a formula.
I don't know where I have done a mistake or something similar.
Pd: Here you have a photo from the book 

Brackets matter. Binary connectives go with outer parentheses, unary connectives and propositional letters don't.
$A \land B$ is not a formula (in the strict sense) because the outer parentheses are missing. $(A \land B$) would be a formula.
$\neg (A)$ and is not a formula because the parentheses are unjustified. (The same would go for $(\neg A)$.) $\neg A$ would be a formula.
Of a different category are $A \Rightarrow B, C$ and $\forall A ...$: They are not formulas because they contain symbols that are not part of the alphabet.
Can you figure out the other ones?
