Are $p \to (q \to r)$ and $p \to (q \wedge r)$ logically equivalent? Is $p \to (q \to r)$ logically equivalent to  $p \to (q \wedge r)$? 
I simplified each one, I got $\neg\, p \vee(q \vee r)$ and $\neg\, p ∨(\neg\, q \wedge r)$ respectively. 
Not sure if my simplification is correct, if not how to simplify it? 
How to find out if I can simplify any further? 
Your advice is greatly appreciated. 
 A: They are not equivalent.
Let $p=1, q=r=0$
Then $q\wedge r = 0$ and so you got
$$p \rightarrow \left(q\wedge r\right) = 0$$
On the other hand $q\rightarrow r = 1$ and thus
$$p \rightarrow \left(q\rightarrow r\right) = 1 $$
A: You can try making a truth table, marking out columns for $p$, $q$, $r$, $q \to r$, and $p \to (q \to r)$, then do another one for $p\to(q\wedge r)$. I'm not going to do it for you but it is pretty straightforward. 
I don't understand what laws you used to simplify the expression. One useful one in this case is the equivalence $a\to b$ with $\neg\,a\wedge b$.    
A: They are not equivalent. Consider the following:
I. If $p$ is prime, then if $q(\neq p)$ is prime, we have $\operatorname{gcd}(p,q)=1$.
II. If $p$ is prime, then $q(\neq p)$ is prime and $\operatorname{gcd}(p,q)=1$.
You can easily see that the above two statements are different-the first one is true and the second is false.
A: The first tells you that IF $p$ is true, AND $q$ is true, then necessarily $r$ is also true. The second tells you that $p$ true implies both $q$ and $r$ are true.  Taking $p$ true, $q$ false, and any value for $r$ satisfies the first but not the second.
So no, they are not equivalent.
A: One way to see this is with the method of analytic tableaux. Experience tells one to first check here whether $\{\neg(p\to (q\to r)),\, \neg (p\to(q \vee r))\}$ is satisfiable (since the tableau is nice and small): if $(p\to (q\to r))\leftrightarrow (p\to(q \vee r))$ is true, then certainly $\neg(p\to (q\to r))$ and $\neg (p\to(q \vee r))$ are true at the same time. But we get
$$\neg(p\to (q\to r)) \\
\neg (p\to(q \vee r)) \\
p \\
\neg(q\to r) \\
p \\
\neg(q\vee r) \\
q \\
\neg r \\
\neg q \\
\neg r\,,$$ which is closed (i.e. it ends in a contradiction). Hence they are not equivalent.
