Entropy of a Finite State Transducer Theorem 7 in Shannon's seminal paper A Mathematical Theory of Communication states:

"The output of a finite state transducer driven by a finite state statistical source is a finite state statistical source, with entropy (per unit time) less than or equal to that of the input. If the transducer is non-singular they are equal."

Basically, it says that a transducer cannot increase the entropy of source. It includes a proof, which I can't make much sense of. I want a justification (even if only by intuition) of how this is true.
Here is the included proof:
Let $\alpha$ represent the state of the source, which produces a sequence of symbols $x_i$; and let $\beta$ be the state of
the transducer, which produces, in its output, blocks of symbols $y_j$. The combined system can be represented by the "product state space" of pairs $(\alpha, \beta)$. Two points in the space $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$, are connected by
a line if $\alpha_1$ can produce an $x$ which changes $\beta_1$ to $\beta_2$, and this line is given the probability of that $x$ in this case. The line is labeled with the block of $y_j$ symbols produced by the transducer. The entropy of the output can be calculated as the weighted sum over the states. If we sum first on $\beta$ each resulting term is less than or equal to the corresponding term for $\alpha$, hence the entropy is not increased. If the transducer is non-singular, let its output be connected to the inverse transducer. If $H'_1$, $H'_2$ and $H'_3$ are the output entropies of the source, the first and second transducers respectively, then $H'_1 \geq  H'_2 \geq H'_3 = H'_1$ and therefore $H'_1 = H'_2$. 
 A: This is known as the data processing inequality and Shannon uses $H'_i$ to mean conditional entropy.
First note that given $Y = f(X)$, the behaviour of $f$ is entirely described by $P(Y|X)$.
Now an explicit statement of the theorem would be to say that, given a Markov chain $X \to Y \to Z$ with factorization $P(X,Y,Z) = P(X)P(Y|X)P(Z|Y)$ we have that $H(X|Y) \le H(X|Z)$.
That is to say that the amount of uncertainty that remains about $X$ after learning $Y$ is never greater than the amount of uncertainty that remains about $X$ after learning $Z$.
This is mainly due to the independence of $X$ and $Z$ given $Y$, which you can see by dividing by $P(Y)$.
$P(X,Z|Y) = \frac{P(X)P(Y|X)P(Z|Y)}{P(Y)} = \frac{P(X,Y)P(Z|Y)}{P(Y)} = P(X|Y)P(Z|Y)$
Of course $P(X,Z|Y) = P(X|Y)P(Z|Y)$ is the definition of independence, which might be better expressed as a property of the mutual information. In our case $I(X,Z|Y) = 0$.
With the definition $I(X;Y) = H(X) - H(X|Y)$, we can rephrase the theorem $H(X|Y) \le H(X|Z)$ using the mutual information:
$I(X;Y) \ge I(X;Z)$
Using the identity $I(X;Y,Z) = I(X;Y) + I(X;Z|Y)$ we find that
$I(X;Y) + I(X;Z|Y) = I(X;Z) + I(X;Y|Z)$
and because we have already established that $I(X;Z|Y) = 0$, this becomes
$I(X;Y) = I(X;Z) + I(X;Y|Z)$
Finally, $I(X;Y|Z) \ge 0$ (a property of mutual information) and we find that
$I(X;Y) \ge I(X;Z)$
which is the original claim.
