Show by using logical connectives laws that $(P\to Q) \land (Q \to R) $ is equivalent to $(P \to R) \land [(P \iff Q) \lor (R \iff Q)]$

I am having trouble with a problem in the book I'm self-studying from. It says the following:

Show that $(P\to Q) \land (Q \to R) $ is equivalent to $(P \to R)$ $\land [(P \iff Q) \lor (R \iff Q)]$ by using logical connectives

I have dedicated so far a hefty amount of time on this problem, and now I'm asking you guys advice/hints or solution as to how to solve this problem. Here is one of the methods I used. Point any flaws that I made.

$(P\to Q) \land (Q \to R) $

(Conditional Law)

$(\neg P \lor Q) \land (\neg Q \lor R) \Rightarrow$

(Distributive Law)

$[(\neg P \land \neg Q)] \lor [Q \land (\neg Q \lor R)] \Rightarrow$

(Distributive Law)

$[(\neg P \land \neg Q) \lor (R \land \neg P)] \lor [(Q \land \neg Q) \lor (R \land Q)] \Rightarrow$

Contradiction

$[(\neg P \land \neg Q) \lor (R \land \neg P)] \lor [(Contradiction) \lor (R \land Q)] \Rightarrow$

(Contradiction Law)

$[(\neg P \land \neg Q) \lor (R \land \neg P)] \lor [ (R \land Q)] \Rightarrow$

Typically at around step five I get stuck or get confused because the problem gets messy. I know you could show it by using the truth-tables. However, the problem says use logical connectives. My questions are: Am I on the right track into solving this problem? Did I make any mistakes? What advice/hints would you give me in my path to solving this problem?

Edit Some of you guys want me to list the laws. Here they are:

DeMorgan's laws

$\neg(P \land Q) \equiv \neg P \lor \neg Q$

$\neg(P \lor Q) \equiv \neg P \land \neg Q$

Commutative laws

$P \lor Q \equiv Q \lor P$

$Q\lor P \equiv P \lor Q$

Associative Laws

$P \land (Q \land R) \equiv (P \land Q) \land R $

$(P \land Q) \land R \equiv P \land (Q \land R) $

Idempotent Laws

$P \land P \equiv P$

$P \lor P \equiv P$

Distributive Laws

$P \land (Q \lor R) \equiv (P \land Q ) \lor (P \land R)$

$P \lor (Q \land R) \equiv (P \lor Q) \land (P \lor R)$

Absorption Laws

$P \lor (P \land Q) \equiv P$

$P \land (P \lor Q) \equiv P$

Tautology Laws

$P \land (tautology) \equiv P$

$P \lor (tautology) \equiv (tautology)$

Contradiction Laws

$P \land (contradiction) \equiv (contradiction)$

$P \lor (contradiction) \equiv P$

Conditional laws

$P \to Q \equiv \neg P \lor Q$

$P \to Q \equiv \neg (P \land \neg Q)$

What we would like to prove is a conjunction, so it suffices to prove each conjunct separately and then glue them together at the end. This problem would probably be easier and more intuitive using proof by contradiction, but after talking with the asker, I will provide a direct proof.

$$\begin{array}{lr} 1. & (P \rightarrow Q)\wedge(Q \rightarrow R) & \text{Premise} \\ 2. & P \rightarrow Q &\text{Simplification, 1}\\ 3. & Q \rightarrow R & \text{Simplification, 1}\\ 4. & \neg{P} \vee Q & \text{Conditional Law, 2}\\ 5. & \neg{Q} \vee R & \text{Conditional Law, 3}\\ 6. & Q \vee \neg{Q} & \text{Tautology} \\ 7. & \neg{P} \vee R & \text{Constructive Dilemma, 4,5,6}\\ 8. & P \rightarrow R & \text{Conditional Law, 7}\\ 9. & (P \rightarrow Q) \vee (Q \rightarrow R) &\text{Addition, 2}\\ 10. & (Q \rightarrow P) \vee (Q \rightarrow R) &\text{Addition, 3}\\ 11. & (P \rightarrow Q) \vee (R \rightarrow Q) &\text{Addition, 2}\\ 12. & Q \vee \neg{Q} & \text{Tautology}\\ 13. & (Q \vee \neg{Q}) \vee (P \vee \neg{R}) & \text{Addidition, 12}\\ 14. & (\neg{Q} \vee P) \vee (\neg{R} \vee Q) & \text{Associative Law, 13}\\ 15. & (Q \rightarrow P) \vee (R \rightarrow Q) & \text{Conditional Law, 14}\\ 16. & \big((P \rightarrow Q) \vee (Q \rightarrow R)\big)\wedge \big((Q \rightarrow P) \vee (Q \rightarrow R)\big) & \text{Conjunction, 9,10}\\ 17. & \big((P \rightarrow Q) \vee (R \rightarrow Q)\big)\wedge \big((Q \rightarrow P) \vee (R \rightarrow Q)\big) & \text{Conjunction, 11,15}\\ 18. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (Q \rightarrow R) & \text{Distributive Law, 16}\\ 19. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (R \rightarrow Q) & \text{Distributive Law, 17}\\ 20. & \Big(\big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (Q \rightarrow R)\Big) \wedge & \\ &\Big(\big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (R \rightarrow Q)\Big) & \text{Conjunction, 18,19}\\ 21. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee \big((Q \rightarrow R) \wedge (R \rightarrow Q) \big) & \text{Distributive Law, 20}\\ 22. & (P \equiv Q) \vee (Q \equiv R) & \text{Definition of Biconditional, 21}\\ \therefore & (P \rightarrow R)\wedge \big((P \equiv Q) \vee (Q \equiv R)\big) & \text{Conjunction, 8,22} \end{array}$$

As desired.