Prove that $(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \implies (P \vee R) \leftrightarrow (Q \vee S).$ Let $P, Q, R$ and $S$ be statements. Prove that
$$(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \implies (P \vee R) \leftrightarrow (Q \vee S).$$
My attempt:
I already check that this logical implication is true, by showing that the conditional statement $(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \to (P \vee R) \leftrightarrow (Q \vee S)$ is a tautology.
Although, I am trying to come up with a derivation of this implication. Here is what I have
$$(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \implies (P \to Q) \wedge (Q \to P) \wedge (R \to S) \wedge (S \to R).$$
Using the Material Implication, I deduce that
$$(P \to Q) \wedge (Q \to P) \wedge (R \to S) \wedge (S \to R)  \implies (\neg P \vee Q) \wedge (\neg Q \vee P) \wedge (\neg R \vee S) \wedge (\neg S \vee R).$$
I thought that maybe I could use the Distributive Law to continue to simplifying this expression. By doing this, I had to rotate my sheet, because the expression was really big.
I can continue this until I got something, but the expression is becoming extremely length.
Is there any hint of what I should do in order to prove this logical implication?
Or any simpler idea of what to do?
Thank you for your attention!
 A: Let $P, Q, R$ and $S$ be statements, and consider the following statement
$$(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \tag{1}$$
Using the rule of inference that states that for any statements $P$ and $Q$ $P \leftrightarrow Q \iff (P \to Q) \wedge (Q \to P),$ we get the following
$$(P \to Q) \wedge (Q \to P) \wedge (R \to S) \wedge (S \to R) \tag{2}$$
Applying the Material Implication to each conditional statement in $(2)$ yields the following expression
$$(\neg P \vee Q) \wedge (\neg Q \vee P) \wedge (\neg R \vee S) \wedge (\neg S \vee R) \tag{3}$$
Using the Rule of inference of Simplification, we get the following four disjunctions
$$\neg P \vee Q \tag{4}$$
$$\neg Q \vee P \tag{5}$$
$$\neg R \vee S \tag{6}$$
$$\neg S \vee R \tag{7}$$
Next, we apply the rule of Addition to each of the previous expressions. Namely, we will add $S$ to $(4),$ $R$ to $(5),$ $Q$ to $(6)$ and $P$ to $(7).$
$$\neg P \vee Q \vee S \tag{8}$$
$$\neg Q \vee P \vee R \tag{9}$$
$$\neg R \vee S \vee Q \tag{10}$$
$$\neg S \vee R \vee P \tag{11}$$
By Adjunction, we get the following expression
$$(\neg P \vee Q \vee S) \wedge (\neg Q \vee P \vee R) \wedge (\neg R \vee S \vee Q) \wedge (\neg S \vee R \vee P) \tag{12}$$
Using the Distributive Law yields that
$$\big( (\neg P \wedge \neg R) \vee (Q \vee S) \big) \wedge \big( (\neg Q \wedge \neg S) \vee (P \vee R) \big) \tag{13}$$
Which, by the De Morgan’s Law, is equivalent to
$$\big( \neg(P \vee R) \vee (Q \vee S) \big) \wedge \big( \neg(Q \vee S) \vee (P \vee R) \big) \tag{14}$$
Using, again, the Material Implication we write our previous expression as
$$\big( (P \vee R) \to (Q \vee S) \big) \wedge \big( (Q \vee S) \to (P \vee R)\big) \tag{15}$$
Finally, using the first rule of inference used here to deduce $(2)$ from $(1),$ we get
$$(P \vee R) \leftrightarrow (Q \vee S) \tag{16}$$
Hence, we have proved that $(1) \implies (16),$ i.e.
$$(P \leftrightarrow Q) \wedge (R \leftrightarrow S) \implies (P \vee R) \leftrightarrow (Q \vee S) \ \text{.  } \text{ } \text{ }\square$$
