Is $(p\rightarrow q)\rightarrow(r\rightarrow s)$ logically equivalent to $(p\rightarrow r)\rightarrow(q\rightarrow s)$? Let $p$, $q$, $r$, and $s$ be any propositions. Then does the following logical equivalence hold?
$$ \big( p \rightarrow q \big) \rightarrow \big( r\rightarrow s \big) \ \equiv \  \big( p \rightarrow r \big) \rightarrow \big( q \rightarrow s \big). \tag{1} $$
My Attempt:

We note that the left-hand-side of (1)  is False if and only if $p \rightarrow q$ is True and $r \rightarrow s$ is False, which in turn holds if and only if
$$
\overline{ \left(  p \equiv T \ \land \ q \equiv F  \right) } \ \land \ \bigl( r \equiv T \  \land \  s \equiv F \bigr),
$$
which is the same as
$$
\bigl( p \equiv F \ \lor \ q \equiv T \bigr) \ \land \ \bigl( r \equiv T \  \land \  s \equiv F \bigr),
$$
that is,
$$
\begin{align} 
& \bigl( p \equiv F \  \land \ r \equiv T \  \land \ s \equiv F \bigr) \\ & \ \ \  \lor \big( q \equiv T \  \land \ r \equiv T \  \land \ s \equiv F \bigr), \end{align} \tag{2}
$$
where $T$ stands for True and $F$ stands for False. Thus the left-hand-side of (1) above is False if and only if (2) holds.


On the other hand, the right-hand-side of (1) is False if and only if $p \rightarrow r$ is True and $q \rightarrow s$ is False, which in turn holds if and only if
$$
 \overline{ \left( p \equiv T \ \land \ r \equiv F  \right) } \ \land \ \bigl( q \equiv T \  \land \  s \equiv F \bigr),
$$
which is the same as
$$
\begin{align} 
& \bigl( p \equiv F \  \lor \ r \equiv T \bigr) \\ 
& \ \ \ \land \ \bigl( q \equiv T \  \land \ s \equiv F \bigr), 
\end{align} 
$$
that is,
$$
\begin{align} 
& \bigl( p \equiv F \  \land \  q \equiv T \  \land \ s \equiv F \bigr) \\ 
& \ \ \  \lor \ \bigl( r \equiv T \  \land \ q \equiv T \ \land \ s \equiv F \bigr), 
\end{align} 
$$
or in other words,
$$
\begin{align} 
& \bigl( p \equiv F \  \land \ q \equiv T \  \land \ s \equiv F \bigr) \\ 
& \ \ \ \lor \  \bigl( q \equiv T \  \land \ r \equiv T \ \land \ s \equiv F \bigr). \end{align}  \tag{3}
$$
Thus the right-hand-side of \label{1} is False if and only if (3) holds.


Now from (2) we find that the left-hand-side of (1) is True if $r \equiv F$ whatever the truth values of $p$, $q$, and $s$ may be, because in that case both the conjunctions that form the components of the disjunction in (2) are False thus making the disjunction in (2) False too. On the other hand, the right-hand-side of (1) is False if
$$ p \equiv F, \qquad q \equiv T, \qquad r \equiv F, \qquad s \equiv F, \tag{4} $$
because for this combination of truth values (3) above holds.


Alternatively, from (3) above we find that the right-hand-side of (1) is True if $q \equiv F$ whatever the truth values of $p$, $r$, and $s$ may be, because in that case both the conjunctions that form the components of the disjunction in (2)  are False thus making the disjunction in (2) False too. On the other hand, the left-hand-side of (1) is False if
$$ p \equiv F, \qquad q \equiv F, \qquad r \equiv T, \qquad s \equiv F, \tag{5} $$
because for this combination of truth values (2) above holds.


Hence the combinations of truth values in (4) and in (5) show that the logical equivalence in (1) above fails to hold.

Am I right?
Is my proof correct, clear, and complete enough? Or, are there any issues of accuracy, clarity, or completeness in this proof?
 A: Here’s another approach, using conjunctive normal form.
We note that
$$
\begin{align} 
 \big(p \rightarrow q \big) \rightarrow \big(r \rightarrow s \big) &\equiv \left( \overline{ p \rightarrow q } \right) \lor \big( r \rightarrow s \big) \\
& \equiv \left( \overline{ \overline{p}  \lor q } \right) \lor \left( \overline{r} \lor s \right) \\
&\equiv \left( \overline{\overline{p}} \land \overline{q} \right) \lor \left( \overline{r}  \lor s \right) \\
&\equiv \left( p \land \overline{q} \right) \lor \left( \overline{r}  \lor s \right) \\
&\equiv \left( p \lor \left( \overline{r}  \lor s \right) \right) \ \land \  \left( \overline{q} \lor \left( \overline{r}  \lor s \right) \right) \\
&\equiv \left( p \lor \overline{r} \lor s \right) \ \land \  \left( \overline{q} \lor \overline{r}  \lor s  \right). 
\end{align} 
$$
Thus we have
$$
 \big(p \rightarrow q \big) \rightarrow \big(r \rightarrow s \big) \ \equiv \ \left( p \lor \overline{r} \lor s \right) \ \land \  \left( \overline{q} \lor \overline{r}  \lor s  \right).  \tag{1} \label1
$$
Interchanging the roles of $q$ and $r$ in \eqref{1} yields
$$
\begin{align} 
 \big(p \rightarrow r \big) \rightarrow \big( q \rightarrow s \big) & \equiv  \left( p \lor \overline{q} \lor s \right) \ \land \  \left( \overline{r} \lor \overline{q}  \lor s  \right) \\ 
&\equiv \left( p \lor \overline{q} \lor s \right) \ \land \  \left( \overline{q} \lor \overline{r}  \lor s  \right).
\end{align} 
$$
Thus we have
$$
 \big(p \rightarrow r \big) \rightarrow \big( q \rightarrow s \big) \ \equiv \ \left( p \lor \overline{q} \lor s \right) \ \land \  \left( \overline{q} \lor \overline{r}  \lor s  \right). \tag{2}\label{2}
$$
From \eqref{1} and \eqref{2}, we note that these two conjunctive normal forms are clearly equivalent when $q$ and $r$ take the same truth values, so a natural next step is to instead test opposite truth values, thus leading to a counterexample.
A: Here is a nice method, called the short or indirect Truth-table method. The basic idea is to try to generate a certain kind of row of a truth-table. ONe of two things will happen: you find that you are able to find such a row, or you find that no such row can exist. Either way, you can draw an appropriate conclusion regarding whatever question you had.
In this case, since we are interested in seeing whether these two statements are equivalent or not, we are interested in the possibility of making the equivalence false ... if we find that we can, then they are not equivalent, but if we find that we cannot, then they are. OK, so here goes:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&&&&&&&\color{red}F\\
\end{array}
Now, there are two ways to try and do this: make the left side true and the right side false, or vice versa. Let's try the first option first, and if that doesn;t work, we can always explore the second. If neither works, then we know the statements are equivalent. Anyway, let's explore the first option:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&&&\color{red}T&&&&F&&&&\color{red}F\\
\end{array}
There is only one way to make a conditional False:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&&&T&&&&F&&\color{red}T&&F&&\color{red}F\\
\end{array}
And again:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&&&T&&&&F&&T&&F&\color{red}T&F&\color{red}F\\
\end{array}
We can now copy the $q$ and $s$ values:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&&\color{red}T&T&&&\color{red}F&F&&T&&F&T&F&F\\
\end{array}
Since $q$ is true, $p \to q$ is true as well:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&\color{red}T&T&T&&&F&F&&T&&F&T&F&F\\
\end{array}
And since $(p \to q) \to (r \to s)$ is True, $r \to s$ must be true:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&T&T&T&&\color{red}T&F&F&&T&&F&T&F&F\\
\end{array}
which forces $r$ to be False:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&T&T&T&\color{red}F&T&F&F&&T&&F&T&F&F\\
\end{array}
copy $r$:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&T&T&T&F&T&F&F&&T&\color{red}F&F&T&F&F\\
\end{array}
forcing $p$:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
&T&T&T&F&T&F&F&\color{red}F&T&F&F&T&F&F\\
\end{array}
copy value of $p$:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
\color{red}F&T&T&T&F&T&F&F&F&T&F&F&T&F&F\\
\end{array}
.... and we've constructed our row!  That is, we have found a truth-value assignment to all variables that sets the equivalence to False: set $p$, $r$, and $s$ to False, and $q$ to True, and that will demonstrate the non-equivalence of the two statements.
Now, we already have our answer to the original question (the statements are not equivalent), but just to show you what you would typically do (amnd how fast this method really is in practice!), let me explore the other option, where this time I use indices to indicate in what order I put down the truth values:
\begin{array}{ccccccccccccccc}
\big( p & \rightarrow & q \big) & \rightarrow & \big( r & \rightarrow & s \big) &  \equiv & \  \big( p & \rightarrow & r \big) & \rightarrow & \big( q & \rightarrow & s \big)\\
\hline
F_{14}&T_4&F_{13}&F_2&T_6&F_5&F_7&F_1&F_{15}&T_{10}&T_8&T_3&F_{12}&T_{11}&F_9\\
\end{array}
So here we get as a counterexample: $p$, $q$, and $s$ are False, and $r$ is True.
And these two counterexamples are exactly the two that you generated, and that other Answers also identify.
A: For the statement you're trying to prove, I tried for a long time to prove/disprove it, but I was unsuccessful. However, try using this tool: https://web.stanford.edu/class/cs103/tools/truth-table-tool/, open two tabs for it, enter (p->q)->(r->s) in the first window, and (p->r)->(q->s) in the second window. You'll see that some values for p,q,r,s lead to different values for (p->q)->(r->s) as compared to (p->r)->(q->s). So this suggests to me that they're not logically equivalent.
A: It is not true in every case. See the truth table:

Source: https://www.erpelstolz.at/gateway/TruthTable.html
