Simplify Logical Equation $\lnot ((P \to (Q \to R)) \to (Q \to (P \to R)))$ I have to simplify (the form with the least atomic propositions):
$$\lnot ((P \Rightarrow (Q \Rightarrow R)) \Rightarrow (Q \Rightarrow (P \Rightarrow R)))$$
Wolfram Alpha generated Truth Table
I have come up with the following before getting stuck:


*

*$$\iff \lnot((P \lor (\lnot Q \lor R)) \Rightarrow (\lnot Q \lor (\lnot P \lor R)))$$ [Implication of v]

*$$\iff \lnot((\lnot P \lor \lnot Q) \lor R)) \Rightarrow ((\lnot Q \lor \lnot P) \lor R)) $$
[Associativity of v]

*$$\iff \lnot(\lnot(P\land Q) \lor R)) \Rightarrow (\lnot (Q \land P) \lor R)) $$
[De Morgan]

*$$\iff \lnot((P \lor Q) \Rightarrow R)) \Rightarrow ((Q \lor P) \Rightarrow  R))   $$
[Implication of v]


Any pointers?
After some feedback:


*

*$$\iff \lnot((P \lor (\lnot Q \lor R)) \Rightarrow (\lnot Q \lor (\lnot P \lor R)))$$ [Implication of v]

*$$\iff \lnot(\lnot(P \lor (\lnot Q \lor R)) \lor (\lnot Q \lor (\lnot P \lor R))) $$ [Implication of v]

*$$\iff \lnot((\lnot(P \lor \lnot Q) \lor R) \lor ((\lnot P \lor \lnot Q) \lor R))$$ [Associativity and commutativity of v]
 A: You can write down and $A\implies B$ as $\neg A\lor B$ and also rewrite $\neg(A\lor B)$ to $\neg A\land \neg B$ and $\neg(A\land B)$ to $\neg A\lor \neg B$. This way you necessarily end up with some expression without "$\implies$" and where all "$\neg$" are right in front of a propositional variable. 
Depending on what you mean with "simplify", you can check if there are some redundancies you can throw away (like $A\land A$ being the same as $A$), and reassemble to implications, if you want. 
In any case, from the truth table, it shows the thing is false in any case, so it's has the same truth value as $A\land \neg A$.
A: $$\lnot \Big((P \Rightarrow (Q \Rightarrow R)) \Rightarrow (Q \Rightarrow (P \Rightarrow R))\Big)\tag 1$$
$$\equiv \lnot\Big(\lnot (P \to (Q\to R)) \lor (Q \to (P \to R))\Big)\tag{(2) implication}$$
$$\equiv (P \to (Q\to R))\land \lnot (Q \to (  P \to R))\tag{(3) DeMorgans}$$
$$\equiv (\lnot P \lor (\lnot Q \lor R)) \land \lnot (\lnot Q \lor (\lnot P\lor R))\tag {(4) implication, $2\times$ }$$
$$\equiv (\lnot P \lor \lnot Q \lor R) \land \lnot (\lnot P \lor \lnot Q \lor R)\tag{(5) associative law}$$
$$\equiv \bot \tag{(6) Complementation}$$

Complementation for any proposition $A,\quad  A \land \lnot A \equiv \bot$, that it $A \land \lnot A$ is a contradidiction, false regardless of truth value assignments to $A$.
A: There's probably a shorter way, but here's a straightforward (admittedly tedious) brute force approach.

The point is that if a statement is false, you can always force it to confess.

\begin{align*}
1.&&&\lnot \Bigl(\bigl(P \rightarrow (Q \rightarrow R)\bigr) \rightarrow \bigl(Q \rightarrow (P \rightarrow R)\bigr)\Bigr)\\[6pt]
2.&&&\lnot\Bigl(\lnot\bigl(P \rightarrow (Q \rightarrow R)\bigr) \lor \bigl(Q \rightarrow (P \rightarrow R)\bigr)\Bigr)\\[6pt]
3.&&&\bigl(P \rightarrow (Q \rightarrow R)\bigr) \land \Bigl(\lnot \bigl(Q \rightarrow (P \rightarrow R)\bigr)\Bigr)\\[6pt]
4.&&&\bigl(\lnot P \lor (Q \rightarrow R)\bigr) \land \Bigl(\lnot \bigl(\lnot Q \lor (P \rightarrow R)\bigr)\Bigr)\\[6pt]
5.&&&\bigl(\lnot P \lor (Q \rightarrow R)\bigr) \land \Bigl(Q \land \bigl(\lnot(P \rightarrow R)\bigr)\Bigr)\\[6pt]
6.&&&\bigl(\lnot P \lor (\lnot Q \lor R)\bigr) \land \Bigl(Q \land \bigl(\lnot(\lnot P \lor R)\bigr)\Bigr)\\[6pt]
7.&&&\bigl(\lnot P \lor (\lnot Q \lor R)\bigr) \land \bigl(Q \land (P \land \lnot R)\bigr)\\[6pt]
8.&&&(\lnot P \lor \lnot Q \lor R) \land (Q \land P \land \lnot R)\\[6pt]
9.&&&\bigl((Q \land P \land \lnot R)\land \lnot P\bigr)\lor \bigl((Q \land P \land \lnot R)\land \lnot Q\bigr)\lor \bigl((Q \land P \land \lnot R)\land R\bigr)\\[6pt]
&&&\qquad\; \vdots\\[6pt]
&&&\bot \lor \bot \lor \bot\\[6pt]
&&&\bot\\[6pt]
\end{align*}
A: $$\lnot ((P \to (Q \to R)) \to (Q \to (P \to R))) \Leftrightarrow \text{ (Implication x 5)}$$
$$(\neg P \lor (\neg Q \lor R)) \land \neg (\neg Q \lor (\neg P \lor R)) \Leftrightarrow \text{ (Association)}$$
$$((\neg P \lor \neg Q) \lor R) \land \neg (\neg Q \lor (\neg P \lor R)) \Leftrightarrow \text{ (Commutation)}$$
$$((\neg Q \lor \neg P) \lor R) \land \neg (\neg Q \lor (\neg P \lor R)) \Leftrightarrow \text{ (Association)}$$
$$(\neg Q \lor (\neg P \lor R)) \land \neg (\neg Q \lor (\neg P \lor R)) \Leftrightarrow \text{ (Complement)}$$
$$\bot$$
A: 
$$\lnot ((P \Rightarrow (Q \Rightarrow R)) \Rightarrow (Q \Rightarrow (P \Rightarrow R)))$$

Well, since by exportation: $(A\to(B\to C)) \iff ((A\wedge B)\to C)$
Or long hand: 

 $$(A\to (B\to C)) ~{\iff (\neg A\vee (\neg B\vee C))\\ \iff ((\neg A\vee \neg B)\vee C)\\ \iff (\neg (A\wedge B)\vee C)\\ \iff (A\wedge B)\to C}$$

Then  $$\lnot \big((P \to (Q \to R)) ~\to~ (Q \to (P \to R))\big)\\\Updownarrow\\ \lnot\big( ((P\wedge Q)\to R)~\to~ ((P\wedge Q)\to R) \big)$$
Which is your stage 4.
And $\neg (X\to X)$ is a contradiction, for any phrase, $X$.  
A: Less tedious and faster way using the following tautological equivalence $$(P \land Q \to R) \leftrightarrow (P\to (Q\to R)  $$
Starting with original statement
$$\lnot ((P \to (Q \to R)) \to (Q \to (P \to R)))$$
$$\lnot ((P \land Q \to R) \to (Q \land P \to R))$$
$$\lnot ((P \land Q \to R) \to (P \land Q \to R))$$
We can see that $A \to A$ is always true, thus $\lnot (A \to A)$ is always false.
