Show that $(P \to R) \land (Q \to R)$ is equivalent to $(P \lor Q) \to R$ I have checked the answer, but do not understand one step.
$(P \to R) \land (Q \to R)$ is equivalent to $(\neg P \lor R) \land (\neg Q \lor R)$
which is equivalent  to $(\neg P \land \neg Q ) \lor R$ 
I don't understand the second equivalent. How to arrive to this?
 A: A different approach for the problem is to convert P,Q,R to Boolean variables and form a truth table where $0$ represents False and $1$ represents True.
\begin{array}{|c|c|} \hline
P & Q & R & P\rightarrow Q & Q\rightarrow R & (P\rightarrow Q)\land(Q \rightarrow R) &P\lor Q &(P \lor Q)\rightarrow R\\ \hline
0 & 0 & 0& 1& 1& 1& 0& 1 \\ \hline
0 & 0 & 1& 1& 1& 1& 0& 1 \\ \hline
0 & 1 & 0& 1& 0& 0& 1& 0 \\ \hline
0 & 1 & 1& 1& 1& 1& 1& 1 \\ \hline
1 & 0 & 0& 0& 1& 0& 1& 0 \\ \hline
1 & 0 & 1& 0& 1& 0& 1& 0 \\ \hline
1 & 1 & 0& 1& 0& 0& 1& 0 \\ \hline
1 & 1 & 1& 1& 1& 1& 1& 1 \\ \hline
\end{array}
So both the statements are equivalent. Hope this helps...
Another approach is to replace $P\rightarrow Q$ by $\neg P \vee Q$ and solve the expression
A: 
$(P \to Q) \land (Q \to R)$ is equivalent to $(\neg P \lor R) \land (\neg Q \lor R)$

No, that is meant to be $(\neg P \lor Q) \land (\neg Q \lor R)$ .  

which is equivalent  to $(\neg P \land \neg Q ) \lor R$ 

$$\begin{align}&(\neg P \lor Q) \land (\neg Q \lor R)
\\=~&((\lnot P\lor Q)\land\lnot Q)\lor((\lnot P\lor Q)\land R)\\=~& (\lnot P\land\lnot Q)\lor((\lnot P\lor Q)\land R)\\=&~ ((\lnot P\land\lnot Q)\lor(\lnot P\lor Q))\land((\lnot P\land\lnot Q)\lor R)\\=&~ (\lnot P\land\lnot Q)\land((\lnot P\land\lnot Q)\lor R)\\=&~ (\lnot P\land\lnot Q)\lor R\end{align}$$
A: One way is to take each pair of expressions and try to reconcile them using known rules, e.g. De Morgan's laws.
The second approach is brute force, namely using truth tables. Not particularly elegant, but it works.
In your case, the unknown variables are P, Q and R. You take 2^3 = 8 combinations of true/false for each variable, and check all your formulas yield the same answer.
You may check this by hand. Example code for Mathematica
formulaA[p_, q_, r_] := (p~Implies~q)~And~(q~Implies~r);
formulaB[p_, q_, r_] := (Not[p]~And~r)~And~(Not[q]~Or~r);
formulaC[p_, q_, r_] := (Not[p]~And~Nor[q])~Or~r;
BooleanTable[{p, q, r} -> formulaA[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaB[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaC[p, q, r], {p, q, r}] // TableForm
It yields something like this. Did you write your first formula correct?
{p,q,r} -> result
{True,True,True}->True
{True,True,False}->False
{True,False,True}->True
{True,False,False}->False
{False,True,True}->True
{False,True,False}->False
{False,False,True}->True
{False,False,False}->True
A: Work with truth values $1,\,0$ instead of true, false, so $P\lor Q=\max\{P,\,Q\}$. Then the claim is that $P,\,Q\le R$ is equivalent to $\max\{P,\,Q\}\le R$, which is trivial.
