Is the following expression true? $(q \Rightarrow p) \Rightarrow(p \Rightarrow q)\equiv(p \Rightarrow q)$ Is the following expression true? $$(q \Rightarrow p) \Rightarrow(p \Rightarrow q)\equiv(p \Rightarrow q)$$
It's actually from a bigger problem, and I'm currently at $(q \Rightarrow p) \Rightarrow(p \Rightarrow q)$. I have to get to $p \Rightarrow q$ to prove the theorem.
$$(q \Rightarrow p) \Rightarrow(p \Rightarrow q)$$
$$(q ∧ p) \Rightarrow(p ∧ q)$$
$$(q ∧ p) ∧ (p ∧ q)$$
$$p∧q$$
$$p \Rightarrow q$$
Is there a faster way to prove this?
Any help is appreciated!
 A: You can just use a truth-table to confirm that this is indeed true.
Or:
$(q \Rightarrow p) \Rightarrow (p \Rightarrow q) \equiv$ (Implication)
$\neg (q \Rightarrow p) \lor (p \Rightarrow q) \equiv$ (Implication x2)
$(q \land \neg p) \lor \neg p \lor q \equiv$ (Absorption)
$\neg p \lor q \equiv$ (implication)
$p \Rightarrow q$
A: I feel compelled to point out that $(q \Rightarrow p) \Rightarrow(p \Rightarrow q)\equiv(p \Rightarrow q)$     is valid in intuitionistic logic (so you don't need to use $a \Rightarrow b \equiv \neg a \lor b$). You also don't need to introduce any connectives (like $\land$ or $\lor$ into the proof). To see this, note that the right-to-left direction of $(q \Rightarrow p) \Rightarrow(p \Rightarrow q)\equiv(p \Rightarrow q)$ is trivial and note that for the left-to-right direction, what you have to do is assume $p$ and $(q \Rightarrow p) \Rightarrow (p \Rightarrow q)$ and from that prove $q$. But that's easy: from the first assumption $p$, infer $q \Rightarrow p$, whence by the second assumption you can infer $p \Rightarrow q$ and then using the assumption $p$ again, you can infer $q$.
A: Yes it is true, but you need to use the fact that $a \Rightarrow b \equiv \neg a \lor b$ as well as the absorption laws (see http://mathworld.wolfram.com/AbsorptionLaw.html) to prove it. Your assumption that $a \Rightarrow b \equiv a \land b$ is not correct.
A: Your work doesn't look correct. My work is below. I use the usual laws of sentential logic, which includes the law saying that $p \to q$ and $\neg p \vee q$ are equivalent (I'll call this the Conditional law). 
$$
\begin{aligned}[t]
(q \to p) \to (p \to q) 
&\equiv \neg(\neg q \vee p) \vee (\neg p \vee q)\\
&\equiv (q \wedge \neg p) \vee (\neg p \vee q) \\
&\equiv [(q \wedge \neg p) \vee \neg p] \vee q \\
&\equiv [\neg p] \vee q \\
&\equiv p \to q
\end{aligned}
\qquad
\begin{aligned}[t]
&{}\\
&\text{DeMorgan; Double Negation} \\
&\text{Associative} \\
&\text{Absorption} \\
&\text{Conditional}
&\end{aligned}
$$
