Logical Proof Simplification I am constructing a formal proof for this set of logical hypotheses:
D,  D ∧ C ⇒ E,  Z ⇒ C ⊢ Z ⇒ E
I have used the deduction theorem / conditional proof to create the following hypotheses, taking Z from the conclusion to be another hypothesis:
H1 = D
H2 = D ∧ C ⇒ E
H3 = Z ⇒ C
H4 = Z
Conclusion = E
I'm trying to be "clever" by simplifying D ∧ C ⇒ E into just C ⇒ E so that I can then use Hypothetical Syllogism rather than modus ponens. Is that a valid simplification? Or can I only simplify D ∧ C when it's on its own (without ⇒ E)? 
E.g. Can a conjunction be simplified when it's implying something else? Or can a conjunction only be simplified when it's on its own?
 A: No!  Here is an example to see why this is an invalid inference:
If I am male and unmarried, then I am a bachelor  (true)
Therefore, if I am male, I am a bachelor (false!)
This is why inference rules for operators can only be applied when that operator is the main operator, e.g. Simplification can only be applied when the main connective is a conjunction (and the Simplification is applied to that very conjunction of course!)
A: If you meant $\bbox[cornsilk, 0.25ex]{D~,~ (D ∧ C) \to E~,~ Z \to C ~\vdash~ Z \to E}$ then this is sound.
$$\begin{array}{l:ll}1 & D & H_1
\\ 2& (D \land C ) \to E & H_2
\\ 3& Z \to C & H_3
\\ \quad 3.1 & Z & A_1
\\ \quad 3.2 & C & 3,3.1,{\to}\mathsf E
\\ \quad 3.3 & D\land C & 1,3.2 , \wedge\mathsf I
\\ \quad 3.4 & E & 2, 3.3, {\to}\mathsf E
\\ 4 & Z\to E & 3.1,3.4,{\to}\mathsf I
\end{array}\\ \therefore D, (D\land C)\to E, Z\to C~\vdash~ Z\to E$$

I'm trying to be "clever" by simplifying D ∧ C ⇒ E into just C ⇒ E so I can use Hypothetical Syllogism rather than modus ponens

No, you cannot simplify.   You are on the wrong side of the conditional to use conjunction elimination.   Instead argue that:
$$D, Z\to C~\vdash~ Z\to(C\land D)$$
Then use HS
