Show that any monotone boolean function can be realize by the following connectives. Let the following be the set of connectives involving {$\top,\ \bot,\ \land,\ \lor$}.
Set $F \leq T$
Show that any monotone Boolean function $f$ can be realize by a wff (well-form formula) using only the above set of connectives.
My attempt is to mimic the prove for the theorem: 
"Let $G$ be an $n$-th place Boolean function, we can find a wff $\alpha$ that realizes $G$"
Attempt:
Case I: ran($f$) = {$F$}, then clearly $f$ can be realize by the constant $\bot$.
Case II: ran($f$) = {$F, T$}.
Mimicking the prove for the theorem, we can look at the $k$ points at which $f$ has the value $T$ and put in a matrix/grid
<$X_{11},...,X_{1n}$ >
....
<$X_{k1},....,X_{kn}$>
For some $k\in \mathbb{N}$
Define $\beta_{ij} = \begin{cases}
      A_{j}, & \text{if}\ X_{ij}=T \\
      A_j\ \lor\ \top, & \text{if}\ X_{ij} = F
    \end{cases}$
Define $\gamma_i = \beta_{i1} \land...\land \ \beta_{in}$
Define $\alpha = \gamma_1 \lor......\lor\ \gamma_k$, $\alpha$ realizes $f$.
Although the proof is similary, I can help but feel that I did something wrong here, mainly becuse I didnt make use of the fact that $f$ is mono-tone.
If I did make a mistake kindly point it out and perhaps point me in the right direction.
 A: It seems to me that your construction should work.
Let $\mathbf{x}_1,\ldots,\mathbf{x}_k$ be such that $f(\mathbf{x}_i) = T$ and $\mathbf{y}_1,\ldots,\mathbf{y}_l$ the remaining vectors such that $f(\mathbf{y}_j) = F$.  We will say that $\mathbf{x}_i = \langle x_{i1},\ldots,x_{in} \rangle$ and similarly for the $\mathbf{y}_j$'s.
To show to your formula realizes $f$, we need to show that it works on all the $\mathbf{x}_i$'s as well as the $\mathbf{y}_j$'s.  It's pretty clear that $\gamma_i(\mathbf{x}_i) = T$ and so $\alpha(\mathbf{x}_i) = T$ for any $i$, as is desired.
To prove correctness w.r.t the $\mathbf{y}_j$'s on the other hand does require monotonicity.  Fix $\mathbf{y}_j$ and let $\mathbf{x}_i$ be arbitrary.  By monotonicity, $\mathbf{x}_i \not\leq \mathbf{y}_j$ and so there is a $p$ such that $x_{ip} = T$ and $y_{jp} = F$.  $A_{p}$ must then appear as a conjunct in $\gamma_i$.  Thus, $\gamma_i(\mathbf{y}_j) = F$, and since $i$ was arbitrary, this holds for every disjunct in $\alpha$.  Thus, $\alpha(\mathbf{y}_j) = F$.
