For Theorem 2, show that$\left(d\right)\Rightarrow \left(a\right),\left(a\right)\Rightarrow \left(c\right),\left(c\right)\Rightarrow \left(b\right),\left(b\right)\Rightarrow \left(c\right)\mathrm{and}\left(c\right)\Rightarrow \left(a\right)$

The theorem (2), describing the line integral of a vector along a closed path, shows that for a closed path $\left(d\right)\Rightarrow \left(a\right)$, $\left(a\right)\Rightarrow \left(c\right)$, $\left(c\right)\Rightarrow \left(b\right)$, $\left(b\right)\Rightarrow \left(c\right)$and $\left(c\right)\Rightarrow \left(a\right)$.

The solenoid fields or divergence less fields are those which have zero divergence, that is $\nabla \cdot F=0$ . If divergence theorem is applied, then surface integral of the function is 0, that is, $\int F\cdot da=0$. It can be expressed as a curl of another vector$F=\nabla \times A$.

As F is expressed as curl of some vector A , that is, $F=\nabla \times A$and divergence of F is 0, that is$\nabla \cdot F=0$, find the curl of function F as,

Substitute $\nabla \times A$ for F into $\nabla .F=0$

$\nabla .(\nabla \times A)=0$

The result is true because the divergence of curl of a vector is always zero. Thus we can say that$d\Rightarrow a$.

According to gauss divergence theorem,

$\iiint \left(\nabla \cdot F\right)dV=\int F\cdot da$

Since the divergence of vector function F is zero, that is$\nabla \cdot F=0$.

Substitute 0 for $\nabla .F$into $\iiint \left(\nabla \cdot F\right)dV=\int F\cdot da$.

$\begin{array}{l}\iiint \left(0\right)dV=\int F\cdot da\\ \int F\cdot da=0\end{array}$

Thus, we can say $a\Rightarrow c$.

As discussed, the surface integral vector F is 0. So$\underset{c}{\overset{}{\int }}F\cdot da=0$ . If the defined surface, have inward surface vector at front surface (1) and outward surface vector at back surface (2). Then, the total surface vector can be expressed as a sum of surface vectors of surface (1) and (2), as,

$\underset{c}{\overset{}{\int }}F\cdot da=\underset{\left(1\right)}{\overset{}{\int }}F\cdot da-\underset{\left(2\right)}{\overset{}{\int }}F\cdot da$

Substitute 0 for$\underset{c}{\overset{}{\int }}F\cdot da$ into$\underset{c}{\overset{}{\int }}F\cdot da=\underset{\left(1\right)}{\overset{}{\int }}F\cdot da-\underset{\left(2\right)}{\overset{}{\int }}F\cdot da$ .

$\begin{array}{l}0=\underset{\left(1\right)}{\overset{}{\int }}F\cdot da-\underset{\left(2\right)}{\overset{}{\int }}F\cdot da\\ \underset{\left(1\right)}{\overset{}{\int }}F\cdot da=\underset{\left(2\right)}{\overset{}{\int }}F\cdot da\end{array}$

As areal vector at both the surface is 0, path from (1) to (2) is independent of surface. Hence, from $c\Rightarrow b$.

As the line integral $\underset{c}{\overset{}{\int }}F\cdot da$ is independent of path from (1) to (2) then for any closed loop, this line integral gives same value as the final and initial points in any closed loop coincide with each other.

Thus, we can write $b\Rightarrow c$.