Velleman exercise 1.5.7a I've been trying to solve the exercise 7(a) of Velleman's "How To Prove It" and haven't succeeded. It asks the verification of the following equivalence:
$$
(P \to Q) \land (Q \to R) = (P \to R) \land ((P \leftrightarrow Q) \lor (R \leftrightarrow Q))
$$
While checking the website for help, I found a question posed by the user "yamad", who, despite his concern with a step futher on the resolution, came to this possible reduced form:
$$(\lnot P \lor Q) \land (\lnot Q \lor R) \land (\lnot P \lor R)$$
The problem is I couldn't even get to this step or any other simplfied form. I would appreciate if someone could provide me a hint.
 A: Let's take the RHS and simplify it a bit:
$$ (P \to R) \land ((P \leftrightarrow Q) \color{red}{\lor} (R \leftrightarrow Q)) \tag{1} $$
Distribute $(P \to R)$ into the parenthesis:
$$(P \to R) \land (P \leftrightarrow Q) \quad \color{red}{\lor} \quad (P \to R) \land (R \leftrightarrow Q) \tag{2} $$
Use the definition of $A \leftrightarrow B \vdash (A \to B) \land (B \to A):$
$$ \color{blue}{(P \to R)} \land (P \to Q) \land \color{blue}{(Q \to P)} \quad \color{red}{\lor} \quad \color{green}{(P \to R)} \land \color{green}{(R \to Q)} \land (Q \to R) \tag{3} $$
At this point we need to assume the result $(A \to B) \land (B \to C) \vdash (A \to C):$
$$ (P \to Q) \land \color{blue}{(Q \to R)} \quad \color{red}{\lor} \quad  \color{green}{(P \to Q)} \land (Q \to R) \tag{4} $$
We know $A \lor A \vdash A:$
$$  (P \to Q) \land (Q \to R) \tag{5} $$

Now we need to prove that
$$(A \to B) \land (B \to C) \vdash (A \to C)$$
which we could verify using a truth table. But the correct proof needs to use  the axioms and theorems of your proof system.
A: The easiest way to verify the identity is to "blast it with an eight-row truth table" as ncmathsadist stated.
Alternatively, you can informally reason this out. If you take the time, you can formalize everything below: 
Suppose $(P \rightarrow Q) \wedge (Q \rightarrow R)$. Then you have $P \rightarrow R$. Now suppose that $\neg((P \leftrightarrow Q) \vee (Q \leftrightarrow R))$. Then (without loss of generality) $P = T$, $Q = F$ and $R = T$. But then $(P \rightarrow Q) \wedge (Q \rightarrow R)$ is false. You can check the other cases. 
Now suppose that $(P \rightarrow R) \wedge ((P \leftrightarrow Q) \vee (Q \leftrightarrow R))$. So suppose you have $(P \rightarrow R)$ and $P \equiv Q$. Then you automatically have $P \rightarrow Q$. You also have $Q \rightarrow P$ and $P \rightarrow Q$. So you have $Q \rightarrow R$. Thus you have $(P \rightarrow Q) \wedge (Q \rightarrow R)$. The other case is done similarly. 
