Is it possible to conclude that the mean of a set is greater than the mean of another set based on means of subsets? Problem 8 of Chapter 2 in Sheldon M. Ross' 5th edition of Introduction to probability and statistics for engineers and scientists is worded as follows:

The sample mean of the weights of the adult women of town A is larger than the sample mean of the weights of the adult women of town B. Moreover, the sample mean of the weights of the adult men of town A is larger than the sample mean of the weights of the adult men of town B. Can we conclude that the sample mean of the weights of the adults of town A is larger than the sample mean of the weights of the adults of town B? Explain your answer.

I tried to describe the problem as precisely as possible in algebraic terms. Here is the description I wrote:
Given:
$$\bar x_{w,A}>\bar x_{w,B} \tag{1}$$
$$\bar x_{m,A}>\bar x_{m,B} \tag{2}$$
Where $$\bar x_{w,A} = \frac{\sum_{i=1}^{n_{w,A}} x_{w,A,i}}{n_{w,A}} \tag{3}$$
$$\bar x_{w,B} = \frac{\sum_{i=1}^{n_{w,B}} x_{w,B,i}}{n_{w,B}} \tag{4}$$
$$\bar x_{m,A} = \frac{\sum_{i=1}^{n_{m,A}} x_{m,A,i}}{n_{m,A}} \tag{5}$$
$$\bar x_{m,B} = \frac{\sum_{i=1}^{n_{m,B}} x_{m,B,i}}{n_{m,B}} \tag{6}$$
Can we conclude that $$\bar x_A > \bar x_B$$?
Where $$\bar x_A = \frac {\sum_{i=1}^{n_{w,A}} x_{w,A,i} + \sum_{i=1}^{n_{m,A}} x_{m,A,i}}{n_{w,A}+n_{m,A}} \tag{7}$$
$$\bar x_B = \frac {\sum_{i=1}^{n_{w,B}} x_{w,B,i} + \sum_{i=1}^{n_{m,B}} x_{m,B,i}}{n_{w,B}+n_{m,B}} \tag{8}$$
My first intuition was to rearrange (3) and (4) into (9) and (10):
$$\bar x_{w, A} n_{w, A} = \sum_{i=1}^{n_{w,A}}x_{w,A,i} \tag {9}$$
$$\bar x_{w, B} n_{w, B} = \sum_{i=1}^{n_{w,B}}x_{w,B,i} \tag {10}$$
Substituting (9) into (7) and (10) into (8) results in the following:
$$\bar x_A = \frac{\bar x_{w,A}n_{w,A} + \bar x_{m,A}n_{m,A}}{n_{w,A}+n_{m,A}}$$
$$\bar x_B = \frac{\bar x_{w,B}n_{w,B} + \bar x_{m,B}n_{m,B}}{n_{w,B}+n_{m,B}}$$
I am basically stuck here. How do I formally show whether we can conclude that $\bar x_A > \bar x_B$? How can I turn this into a formal mathematical proof?
 A: We cannot in general say anything about a trend in the population based on trends within subgroups of the population, and often the reverse of what we would expect is true. This phenomenon is called Simpson's Paradox and it is well documented in real world experiments, in particular for gender-based partitions of the population.
A: If women on the whole weigh less than men and town $A$ has a higher proportion of women the mean adult weight can be lower in town $A$.  For a simple example, assume A has three women weighing $1.1$ and one man weighing $2.1$ while B has one woman weighing $1$ and one man weighing $2$.  The mean adult in $A$ weighs $1.35$ while the mean adult in $B$ weighs $1.5$
A: I scoured through the internet and books on mathematical proof to build up a very rudimentary idea of how to construct a basic mathematical statement and evaluate whether it is true or not. Rephrasing the proposition as a conditional statement helps a lot here, namely:
$$\boxed{\text{IF } (\bar x_{w,A}>\bar x_{w,B} \text{ AND } \bar x_{m,A} > \bar x_{m,B}), \text{THEN } (\bar x_A > \bar x_B)}$$
Where the hypothesis $P$ is ($\bar x_{w,A}>\bar x_{w,B}$ and $\bar x_{m,A} > \bar x_{m,B}$) and the conclusion $Q$ is ($\bar x_A > \bar x_B$). As mentioned by @Ross Millikan showing a statement is false basically boils down to finding a counter example. The following table outlines all the possibilities in the context of a conditional statement:
\begin{array} {|r|r|}\hline P & Q & P \implies Q \\ \hline T & T & T \\ \hline T & F & F \\ \hline F & T & T \\ \hline F & F & T \\ \hline  \end{array}
Disproving the entire conditional statement would put us in row 2 of the above table. That is the only situation where $P \implies Q$ is false. The whole conditional statement being false implies that $Q$ is false. If I had to write this using my basic notions of this proof notation I would write $ \neg (P \implies Q) \implies \neg Q $. In English:

The negation of P implies Q itself implies the negation of Q, or that Q is false

Therefore, finding a counterexample to $\text{IF } (\bar x_{w,A}>\bar x_{w,B} \text{ AND } \bar x_{m,A} > \bar x_{m,B}), \text{THEN } (\bar x_A > \bar x_B)$ itself implies that $(\bar x_A > \bar x_B)$ is false.
The other formulas in my initial question are basically definitions. They are assumed to be true in the context of the conditional statement and no proof of them is sought as far as I understand.
