Trying to find a prespective on a truth table I'm currently reading "A Book of Set Theory" by Charles C. Pinter and I'm a bit stuck in the first chapter Exerciser 1.1, problem 7. Which states:

Prove that for all sentences $P, Q, R, S$, if $P \implies Q$ and $R \implies S$, then
a) $P \vee R \implies Q \vee S$
b) $P \wedge R \implies Q \wedge S$

In the same set of exercises the author had me solve the problems using truth tables which I did. Here is the the truth table for a:
$$\begin{array}{c|c|c|c|} 
P, Q, R, S & P \implies Q & R \implies S & P \vee R \implies Q \vee S \\ \hline
\text{T, T, T, T} & \text{T} & \text{T} & \text{T} \\ \hline
\text{T, T, T, F} & \text{T} & \text{F} & \text{T} \\ \hline
\text{T, T, F, T} & \text{T} & \text{T} & \text{T} \\ \hline
\text{T, T, F, F} & \text{T} & \text{T} & \text{T} \\ \hline
\text{T, F, T, T} & \text{F} & \text{T} & \text{T} \\ \hline
\text{T, F, F, T} & \text{F} & \text{T} & \text{T} \\ \hline
\text{T, F, T, F} & \text{F} & \text{F} & \text{F} \\ \hline
\text{T, F, F, F} & \text{F} & \text{T} & \text{F} \\ \hline
\text{F, T, T, T} & \text{T} & \text{T} & \text{T} \\ \hline
\text{F, T, T, F} & \text{T} & \text{F} & \text{T} \\ \hline
\text{F, T, F, T} & \text{T} & \text{T} & \text{T} \\ \hline
\text{F, F, T, T} & \text{T} & \text{T} & \text{T} \\ \hline
\text{F, F, F, T} & \text{T} & \text{T} & \text{T} \\ \hline
\text{F, F, T, F} & \text{F} & \text{F} & \text{F} \\ \hline
\text{F, T, F, F} & \text{T} & \text{T} & \text{T} \\ \hline
\text{F, F, F, F} & \text{T} & \text{T} & \text{T} \\ \hline
\end{array}$$
Now I'm trying to make sense of what I'm seen. How do I know if this is valid? If we interpret the original question as $(P \implies Q) \wedge (R \implies S) \equiv P \vee R \implies Q \vee S$. Then we can't prove it because we have cases where $P \implies Q$ is Truth and $R \implies S$ is False and yet we see that $P \vee R \implies Q \vee S$ is Truth.
Assuming that the question expects that the statements are indeed true. Is there something I'm missing? Also could this be proven using inference? If so how could it be done? As far as I got was that it resembles to the Constructive Dilemma (CD). Any help would be greatly appreciated. :)
 A: The problem does not ask you to prove an equivalence, but an implication.  Whenever $P \Rightarrow Q$ and $R \Rightarrow S$, then $P \vee R \Rightarrow Q \vee S$.  This is confirmed by your truth table.  The converse does not need to hold, and in fact it doesn't.
A: Just for the sake of completion. I was able to solve it using rules of inference. To get $P \lor R \implies Q \lor S$ I did the following:


*

*$P \implies Q$ (Premise)

*$R \implies S$ (Premise)

*$(P \implies Q) \lor S$ (1, Addition)

*$(R \implies S) \lor Q$ (2, Addition)

*$(\neg{P} \lor Q) \lor S$ (3, Switcheroo)

*$(\neg{R} \lor S) \lor Q$ (4, Switcheroo)

*$\neg{P} \lor (Q \lor S)$ (5, Association)

*$\neg{R} \lor (Q \lor S)$ (6, Association/Commutative)

*$[\neg{P} \lor (Q \lor S)]\land[\neg{R} \lor (Q \lor S)]$ (7,8, Conjugation)

*$(Q \lor S) \lor (\neg{P} \land \neg{R})$ (9, Distribution)

*$(Q \lor S) \lor \neg{(P \lor R)}$ (10, De Morgan)

*$\neg{(Q \lor S)} \implies \neg{(P \lor R)}$ (11, Switcheroo)

*$(P \lor R) \implies (Q \lor S)$ (12, Contrapositive)


For $(P \land R) \implies (Q \land S)$ I did the following:


*

*$P \implies Q$ (Premise)

*$R \implies S$ (Premise)

*$(P \implies Q) \lor \neg{R}$ (1, Addition)

*$(R \implies S) \lor \neg{P}$ (2, Addition)

*$(\neg{P} \lor Q) \lor \neg{R}$ (3, Switcheroo)

*$(\neg{R} \lor S) \lor \neg{P}$ (4, Switcheroo)

*$(\neg{P} \lor \neg{R}) \lor Q$ (5, Association/Commutative)

*$(\neg{P} \lor \neg{R}) \lor S$ (6, Association/Commutative)

*$[(\neg{P} \lor \neg{R}) \lor Q] \land [\neg{R} \lor \neg{P}) \lor S]$ (7,8, Conjunction)

*$(\neg{P} \lor \neg{R}) \lor (Q \land S)$ (9, Distribution)

*$\neg{(P \land R)} \lor (Q \land S)$ (10, De Morgan)

*$(P \land R) \implies (Q \land S)$ (11, Switcheroo)


Took me a while to answer this, since I had to read up on logical inference. Good link too http://www.zweigmedia.com/RealWorld/logic/logicintro.html
