Question about Wronskian. I'm studying Linear Homogeneous Equations and Wronskian, and I saw Theorems that

Let $S$ be the set of all solutions of $y^{''}$ + $py^{'}$ + $qy$ = $0$, where $p(t)$ and $q(t)$ are continuous in an open interval I, and $w[y_1, y_2]$ form Wronskian of $y_1$ and $y_2$ . Then TFAE:
I.   <$y_1$, $y_2$> = $S$  (< > means span).
II.  $w[y_1, y_2](t_o) \neq 0$ $\,\forall t_0$ in $\,I$
III. $w[y_1, y_2](t_o) \neq 0$ for some $t_0$ in $I$
IV.  $y_1$ and $y_2$ are linearly independent

How can I prove them? I heard we should prove them in sequence, like I -> II -> III -> IV, but I have no idea how to prove I -> II. Can anyone help me?
 A: With
$w[y_1, y_2] = \det \left ( \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end {bmatrix} \right ) = y_1y_2' - y_2y_1', \tag 1$
we find
$w'[y_1, y_2] = (y_1y_2' - y_2y_1')'$
$= y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2y_1'' = y_1y_2'' - y_1''y_2; \tag 2$
if we now write the equation
$y'' + py' + qy = 0 \tag 3$
in the form
$y'' = -py' - qy, \tag 4$
we have
$y_i'' = -py_i' - qy_i, \; i = 1, 2, \tag 5$
which we may substitute into (2), yielding
$w'[y_1, y_2] = y_1y_2'' - y_1''y_2$
$= y_1(-py_2' - qy_2) - y_2 ( -py_1' - qy_1)$
$= -py_1y_2' - qy_1y_2 + py_1'y_2 + qy_1y_2$
$= p(y_1'y_2 - y_1y_2') = -pw[y_1, y_2]; \tag 6$
we thus see that $w[y_1, y_2]$ satisfies the ordinary differential equation
$w'[y_1, y_2] = -pw[y_1, y_2]; \tag 7$
the unique solution to this equation taking the value $w[y_1, y_2](t_0)$ at $t = t_0$ is readily seen to be
$w[y_1, y_2](t) = w[y_1, y_2](t_0) \exp \left ( -\displaystyle \int_{t_0}^t p(s) \; ds \right ). \tag 8$ 
Since
$\exp \left (- \displaystyle \int_{t_0}^t p(s) \; ds \right ) \ne 0, \; \forall t \in I, \tag 9$
we deduce that
$w[y_1, y_2](t) \ne 0, \forall t \in I \Longleftrightarrow \exists t_0 \in I, \; w[y_1, y_2](t_0) \ne 0. \tag{10}$
(10) is in fact an expression of the equivalence of points (II) and (III); to see that these imply (IV), we assume that that $y_1$ and $y_2$ are in fact linearly dependent; then there exist
$a, b \in \Bbb R, \tag{11}$
not both $0$, with
$ay_1 + by_2 = 0; \tag{12}$
then
$ay_1' + by_2' = 0, \tag{13}$
and the columns of the matrix
$\begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end {bmatrix} \tag{14}$
are linearly dependent; but this implies
$w[y_1, y_2](t) = 0, \; \forall t \in I, \tag{15}$
contradicting (II), (III), and thus the linear independence of $y_1$, $y_2$ is established; likewise, if we assume (IV), then the columns of the matrix (14) must be linearly independent, and hence
$w[y_1, y_2](t) \ne 0, \forall t \in I; \tag{16}$
thus (II) and (III) bind.
Finally, the set $S$ of all solutions to (3) is well-known to be a two-dimensional vector space over $\Bbb R$; clearly
$\langle y_1, y_2 \rangle \subset S; \tag{17}$
since $y_1$ and $y_2$ are linearly independent, 
$\dim \langle y_1, y_2 \rangle = 2 = \dim S, \tag{18}$
from which it follows that
$\langle y_1, y_2 \rangle = S, \tag{19}$
which is (I); likewise, (I) implies that $\langle y_1, y_2 \rangle$ are linearly independent, lest
$\dim \langle y_1, y_2 \rangle < 2 = \dim S, \tag{20}$
and 
$\langle y_1, y_2 \rangle = S \tag{21}$
is impossible since the dimensions don't agree.
Thus we see that
$[(II) \equiv (III)] \equiv (IV) \equiv (I), \tag{22}$
i.e., that
$(I) \equiv (II) \equiv (III) \equiv (IV), \tag{23}$
$OE\Delta$.
