Rudin's PMA: theorem 10.43 
This is the definition which we need for the proof of the theorem :

This is the definition of the $\mathscr C''$ - equivalent :

There is the theorem:
Suppose E is an open set in $R^3$, $u$ $\in$ $\mathscr C''(E)$, and $G$ is a vector field in $E$, of class $C''$ .
$(a)$ if $F$ $=$ $\nabla$$u$, then $\nabla$ $\times$ $F$ $=$ $0$ .
$(b)$ if $F$ $=$ $\nabla$ $\times$ $G$, then $\nabla$ $\cdot$ $F$ $=$ $0$.
Furthermore, if $E$ is $\mathscr C''$-equivalent to a convex set, then $(a)$ and $(b)$ have converses, in which we assume that $F$ is a vector field in $E$, of class $\mathscr C'$:
$(a')$ if $\nabla$ $\times$ $F$ $=$ $0$, then $F$ = $\nabla$$u$ for some $u$ $\in$ $\mathscr C''(E)$.
$(b')$ if $\nabla$ $\cdot$ $F$ $=$ $0$, then $F$ $=$ $\nabla$ $\times$ $G$ for some vector field $G$ in $E$, of class $\mathscr C''$
There is the proof  :
If we compare the definitions of $\nabla$$u$, $\nabla$ $\times$ $F$, and $\nabla$ $\cdot$ $F$ with the differential forms $\lambda_F$ and $\omega_F$ given by ($124$) and ($125$), we obtain the following four statements:
$F$ $=$ $\nabla$$u$   if and only if $\lambda_F$ $=$ $du$.                    ($\star$).
$\nabla$ $\times$ $F$ $=$ $0$ if and only if $d\lambda_F$ $=$ $0$.             ($\ast$)
$F$ $=$ $\nabla$ $\times$ $G$   if and only if $\omega_F$ $=$ $d\lambda_G$.  ($\oplus$)
$\nabla$ $\cdot$ $F$ $=$ $0$   if and  only if $d\omega_F$ $=$ $0$.            ($\circ$)
I could n't understand these four statements ($\star$),($\ast$),($\oplus$),($\circ$).
Any help would be appreciated.
 A: For the first one, $F=\nabla u$ iff $F_i=D_iu, i=1,2,3$ where $D_iu$ is the partial derivative of $u$ w.r.t. the $i$-th variable. Since $du=(D_1u)\,dx+(D_2u)\,dy+(D_3u)\,dz$, we have $\lambda_F=du$ is equivalent to $F_i=D_iu$ for all $i$ and hence equivalent to $F=\nabla u$.
For the second one, we have
\begin{align*}d\lambda_F=&dF_1\wedge dx+dF_2\wedge dy+dF_3\wedge dz\\
=&(D_1F_1\,dx+D_2F_1\,dy+D_3F_1\,dz)\wedge dx+(D_1F_2\,dx+D_2F_2\,dy+D_3F_2\,dz)\wedge dy\\&+(D_1F_3\,dx+D_2F_3\,dy+D_3F_3\,dz)\wedge dz
\end{align*}
where $dx\wedge dx=0$ and $dx\wedge dy=-dy\wedge dx$. Hence the above is equal to (after reorganizing terms)
$$d\lambda_F=(D_1F_2-D_2F_1)dx\wedge dy+(D_1F_3-D_3F_1)dx\wedge dz+(D_2F_3-D_3F_2)dy\wedge dz$$
Comparing this with $\nabla\times F$, we get the second statement.
For the third one, it is very similar to the second one.
For the last one,
\begin{align*}
d\omega_F&=dF_1\wedge dy\wedge dz+dF_2\wedge dz\wedge dx+dF_3\wedge dx\wedge dy\\
&=D_1F_1dx\wedge dy\wedge dz+D_2F_2dy\wedge dz\wedge dx+D_3F_3dz\wedge dx\wedge dy\\
&=(D_1F_1+D_2F_2+D_3F_3)dx\wedge dy\wedge dz
\end{align*}
Hence $d\omega_F=0$ iff $\nabla\cdot F=0$.
A: The statements themselves are reasonable. Rudin proceeds:
$F=\nabla u \implies \lambda_F=du \implies d\lambda_F= 0 \implies \nabla \times F=0$.
$\nabla \times F=0 \implies d\lambda_F= 0 \implies \lambda_F=du \implies F=\nabla u $.
to prove statements (a) and (a').
$F=\nabla\times G \implies\omega_F=d\lambda_G\implies d\omega_F=0\implies \nabla\cdot F=0$
$\nabla\cdot F=0\implies d\omega_F=0 \implies \omega_F=d\lambda_G\implies F=\nabla\times G$
to prove statements (b) and (b').
All of the central 'implies' are justified using previous theorems in the book.
