If the hypothetical syllogism is a theorem, then:
$$(p\overline{q})+(q\overline{r}) + \overline{p} + r = 1.$$
Here is one way of demonstrating that:
$$\overline{(p\overline{q})+(q\overline{r}) + \overline{p} + r} = 0.$$
$$(\overline{p} + q)(\overline{q} + r) p \overline{r} = 0.$$
$$p (\overline{p} + q)\overline{r}(\overline{q} + r) = 0.$$
$$(p\overline{p} + pq)(\overline{r}\overline{q} + \overline{r}r) = 0.$$
$$(0 + pq)(\overline{r}\overline{q} + 0) = 0.$$
$$(pq)(\overline{r}\overline{q}) = 0.$$
$$(p\overline{r})(q\overline{q}) = 0.$$
$$(p\overline{r})(0) = 0.$$
$$0 = 0.$$
I didn't read the comments, so I suspect you already got a satisfactory answer, but if not, this might be of some help. If some step is wrong or unclear, leave a comment and we'll find the appropriate rule of boolean algebra that justifies it.
Since you explicitly asked for a direct proof and I am stuck at an airport, I'll add the following:
$$(p\overline{q})+(q\overline{r}) + \overline{p} + r = 1.$$
$$(~(p\overline{q})+ \overline{p}~) + (~(q\overline{r}) + r~) = 1.$$
$$(~\overline{q} + \overline{p}~) + (~q + r~) = 1.$$
$$(~\overline{q} + q~) + (~\overline{p} + r~) = 1.$$
$$(1) + (~\overline{p} + r~) = 1.$$
$$1 + \dots = 1.$$
A boolean function is simply a function on boolean variables that produces a boolean result. There is no requirement that it must built in a certain way.
However, if you insist that you want some chain of logical operators, for any fixed $m,k$ we can build such an $f$. Suppose for now that $m \ge k$.
Note that $x$ has at least as many active values as $y$, if and only if there is a way to match up the elements of $x$ with $y$ so that every active element of $y$ is matched with an active element of $x$. That is, if $x_i$ and $y_j$ are matched, then either $y_j = 0$ or $x_i = 1$, or in symbols, $y_j \implies x_i$.
To formalize this, define a pairing to be a set $P \subset \{1, \ldots, m\} \times \{1, \ldots k\}$ such if $(i, j), (r,s) \in P$, then $i = r \iff j = s$, and for all $1 \le j \le k$, there exists some $i$ such that $(i,j) \in P$. Let $\scr P$ be the set of all such pairings. If $p \in P \in \mathscr P$, denote its coordinates by $p = (i_p, j_p)$.
So $x$ has at least as many active elements as $y$ if and only if there exists a $P\in \scr P$ such that
$$\bigwedge_{p\in P} (y_{j_p} \implies x_{i_p})$$
That is, when
$$\bigvee_{P \in \mathscr P}\bigwedge_{p\in P} (y_{j_p} \implies x_{i_p})$$
is true. So $$f(x,y) = \bigvee_{P \in \mathscr P}\bigwedge_{p\in P} (y_{j_p} \implies x_{i_p})$$
When $m < k$, there are no pairings. Instead define a partial pairing as a tuple $(P, Q)$ where $P$ is a pairing of some $m$ elements of $\{1,\ldots, k\}$ with $\{0,\ldots, m\}$, and $Q$ is the set of $k-m$ indices that were not paired. Let $\scr Q$ be the set of partial pairings.
Now we can express
$$f(x,y) = \bigvee_{(P,Q) \in \mathscr Q}\left(\bigwedge_{p\in P} (y_{j_p} \implies x_{i_p})\wedge \bigwedge_{j \in Q} (\lnot y_j)\right)$$
<Edit>
The above only encodes $x$ having as many active values as $y$. To also encode that at least one value must be active, we have
$$f(x,y) = \left(\bigvee_{i=1}^m x_i\right) \wedge \left(\bigvee_{(P,Q) \in \mathscr Q}\left(\bigwedge_{p\in P} (y_{j_p} \implies x_{i_p})\wedge \bigwedge_{j \in Q} (\lnot y_j)\right)\right)$$
</Edit>
All this bit about pairings is just a way of expressing the short-hand notation that holds for every $m,k$. For a particular $m,n$, it is just a matter of listing every every possible pairing. For example, when $m = k = 2$, this is
$$f(x, y) = \left(x_1 \vee x_2\right) \wedge \left(\left[ (y_1 \implies x_1) \wedge (y_2 \implies x_2) \right] \vee \left[(y_1 \implies x_2) \wedge (y_2 \implies x_1) \right]\right)$$
but as $m, k$ get larger, this quickly grows into a ridiculously long expression. But despite that, it still accurately represents $f(x,y)$.
There are undoubtedly simpler expressions that are equivalent. But what this shows is that it is always possible to express $f(x,y)$ in terms of logical operators. Which is something that is true of any boolean function.
<Edit> Adding a few simple examples.
To simplify the functions, I'm just doing "$x$ has at least as many active elements as $y$". To get the actual condition of the question, "and" the functions with $(x_1 \vee x_2 \vee \ldots \vee x_m)$.
If we match a set $a$ of three variables with set $b$ of two, there are $6$ possible pairings between them, each with one member of $a$ left over:
$$\begin{array}{c|ccc}
& a_1 & a_2 & a_3\\
\hline
p_1 & b_1 & b_2 & - \\
p_2 & b_2 & b_1 & - \\
p_3 & b_1 & - & b_2 \\
p_4 & b_2 & - & b_1 \\
p_5 & - & b_1 & b_2 \\
p_6 & - & b_2 & b_1
\end{array}$$
If $a = x$ and $b = y$, then each pairing is represented by the expressions
$$\begin{array}{c|c}
p_1 & (y_1 \implies x_1) \wedge (y_2 \implies x_2) \\
p_2 & (y_2 \implies x_1) \wedge (y_1 \implies x_2) \\
p_3 & (y_1 \implies x_1) \wedge (y_2 \implies x_3) \\
p_4 & (y_2 \implies x_1) \wedge (y_1 \implies x_3) \\
p_5 & (y_1 \implies x_2) \wedge (y_2 \implies x_3) \\
p_6 & (y_2 \implies x_2) \wedge (y_1 \implies x_3)
\end{array}$$
When $x$ has the extra elements, it doesn't matter if they are active or not, so nothing has to be added for them.
When $x$ has more or equal active elements, at least one of these statements will be true, and vice versa. So the total condition is
$$\begin{align}f(x,y) =\ &[(y_1 \implies x_1) \wedge (y_2 \implies x_2)]\ \vee \\
&[(y_2 \implies x_1) \wedge (y_1 \implies x_2)]\ \vee \\
&[(y_1 \implies x_1) \wedge (y_2 \implies x_3)]\ \vee \\
&[(y_2 \implies x_1) \wedge (y_1 \implies x_3)]\ \vee \\
&[(y_1 \implies x_2) \wedge (y_2 \implies x_3)]\ \vee \\
&[(y_2 \implies x_2) \wedge (y_1 \implies x_3)]\end{align}$$
If $a = y, b = x$, so it is $y$ that has the unmatched elements, then it is necessary to guarantee that the unmatched elements are not active:
$$\begin{array}{c|c}
p_1 & (y_1 \implies x_1) \wedge (y_2 \implies x_2) \wedge \lnot y_3 \\
p_2 & (y_1 \implies x_2) \wedge (y_2 \implies x_1) \wedge \lnot y_3 \\
p_3 & (y_1 \implies x_1) \wedge \lnot y_2 \wedge (y_3 \implies x_2) \\
p_4 & (y_1 \implies x_2) \wedge \lnot y_2 \wedge (y_3 \implies x_1) \\
p_5 & \lnot y_1 \wedge (y_2 \implies x_1) \wedge (y_3 \implies x_2) \\
p_6 & \lnot y_1 \wedge (y_2 \implies x_2) \wedge (y_3 \implies x_1)
\end{array}$$
And again, we "or" the various pairings together to get the full function:
$$\begin{align}f(x,y) =\ &[(y_1 \implies x_1) \wedge (y_2 \implies x_2) \wedge \lnot y_3]\ \vee \\
&[(y_1 \implies x_2) \wedge (y_2 \implies x_1) \wedge \lnot y_3]\ \vee \\
&[(y_1 \implies x_1) \wedge \lnot y_2 \wedge (y_3 \implies x_2)]\ \vee \\
&[(y_1 \implies x_2) \wedge \lnot y_2 \wedge (y_3 \implies x_1)]\ \vee \\
&[\lnot y_1 \wedge (y_2 \implies x_1) \wedge (y_3 \implies x_2)]\ \vee \\
&[\lnot y_1 \wedge (y_2 \implies x_2) \wedge (y_3 \implies x_1)]\end{align}$$
<\Edit>
Best Answer
Yes, this one is easy enough to analyze without a 'brute force' truth table.
Let's just call the inputs $1$, $2$, $3$, $4$, $5$.
First, it is easy to see that the value of $5$ will never make a difference as to whether the threshold gets passed or not, so $5$ is a "don't care": we can completely ignore it in our analysis and hence it need not show up in our boolean expression.
Second, focusing just on $1$, $2$, $3$, $4$, you can quickly observe that the threshold will only be passed if at least one of $1$ or $2$ are 'on'.
Now, if $1$ is on but $2$ is not, then the threshold will be passed as long as neither $3$ nor $4$ are on. This gives us the term $1 \land \neg 2 \land \neg 3 \land \neg 4$. But actually, it's ok if $2$ would be on as well. So, this term can be simplified to just $1 \land \neg 3 \land \neg 4$
If $2$ is on but $1$ is not, then the threshold is passed as long as not both $3$ and $4$ are on. This gives us $\neg 1 \land 2 \land \neg (3 \land 4)$. And again, since it's actually ok for $1$ to be on, we get $2 \land \neg (3 \land 4)$
Finally, if both $1$ and $2$ are on, then the threshold will be passed no matter what. So: $1 \land 2$
Since these are the only 3 situations in which the threshold will be passed, we thus get:
$$(1 \land \neg 3 \land \neg 4) \lor (2 \land \neg (3 \land 4)) \lor (1 \land 2)$$