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