Show that the dual of $\min c^{T}x+d^{T}x^{'}$ and $\max a^{T}x+b^{T}x^{'}$ are equivalent Show that $(1)$ can be written in the form $(2)$
$(1)$
$\min c^{T}x+d^{T}x^{'}$
$\operatorname{s.t.}$
$Ax+Bx^{'}\geq a$
$Cx+Dx^{'}= b$
where $x, x^{'} \geq 0$ and 
and 
$(2)$ $\max a^{T}y+b^{T}y^{'}$
$\operatorname{s.t.}$
$A^{T}y+C^{T}y^{'}\leq c$
$By+Dy^{'}= d$
where $y,y^{'} \geq 0$
My idea:
The restrictions $(1)$ can be written in the form:
$\mathcal{P}(\begin{pmatrix}
A & -B \\
C & D \\
-C &-D
\end{pmatrix},\begin{pmatrix} -a \\ b \\ -b \end{pmatrix})$ 
Then the dual LP to $(1)$ can be written as:
$\max \begin{pmatrix} a & -b & b \end{pmatrix}^{T}(y, -y)$
$\operatorname{s.t.}$
$\begin{pmatrix}
A & -B \\
C & D \\
-C &-D
\end{pmatrix}^{T}(y,y^{'})=-\begin{pmatrix} c \\ d \end{pmatrix}$
But how do I get to writing this in the form of $(2)$
Any help is greatly appreciated.
 A: Primal
min   $c^{T}x + d^{T}x'$
s.t
$Ax+  Bx' \geq a$
$Cx + Dx' = b$
$x \geq 0$
$x' \geq 0$
and we can rewrite de primal problem
min   $c^{T}x + d^{T}x'$
s.t
$Ax+  Bx' \geq a$
$Cx + Dx' \geq b$
$-Cx - Dx' \geq -b$
$x \geq 0$
$x' \geq 0$
Dual
max   $a^{T}y +b^{T}y - b^{T}y''$
s.t
$A^{T}y +C^{T}y' - C^{T}y'' \leq c$
$B^{T}y +D^{T}y' - D^{T}y'' \leq d$
$y \geq 0$
$y' \geq 0$
$y'' \geq 0$
and we can rewrite the dual problem
$z = y' - y''$
max  $a^{T}y + b^{T}z$
s.t
$A^{T}y +C^{T}z \leq c$
$B^{T}y +D^{T}z \leq d$
$y \geq 0$
z unrestricted
A: Your problem $(1)$ consists of minimizing $c^Tx + d^Tx^\prime$ under the conditions
\begin{equation}
Ax + Bx^\prime \geq a \qquad (i.) \\
Cx + Dx^\prime = b \qquad (ii.)
\end{equation}
Suppose $z = \min c^Tx + d^Tx^\prime$. I want to find a good valid lower bound for $z$. 
If I multiply $(i.)$ by a posivite value $y$ and $(ii.)$ by a rational value $y^\prime$, I get
\begin{equation}
Axy + Bx^\prime y \geq ay \qquad (i^\prime.) \\
Cxy^\prime + Dx^\prime y^\prime = by^\prime \qquad (ii^\prime.)
\end{equation}
Summing up $(i^\prime.)$ and $(ii^\prime.)$ one has
\begin{equation}
x(Ay + Cy^\prime) + x^\prime(By + Dy^\prime) \geq ay +  by^\prime 
\end{equation}
Notice that 
\begin{equation}
 c^Tx + d^Tx^\prime \geq x(Ay + Cy^\prime) + x^\prime(By + Dy^\prime) \geq ay +  by^\prime 
\end{equation}
holds if and only if
\begin{equation}
Ay + Cy^\prime \leq c \qquad (iii.)\\
By + Dy^\prime \leq d \qquad (iv.)
\end{equation}
Therefore, the best lower bound for z is obtained by maximizing $ay +  by^\prime$ such that $(iii.)$ and $(iv.)$ hold.
PS.: I did not worry of transposing the matrix at the right places but you can do it on your own and see that the reasoning holds.
