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$$ \begin{align} &(p \to (q \to r)) \to ((p \wedge q) \to r) && \\ \iff &\neg (\neg p \vee (\neg q \vee r)) \vee (\neg(p \wedge q) \vee r) && \text{expression for implications} \\ \iff &(p \wedge q \wedge \neg r) \vee (\neg p \vee \neg q \vee r) && \text{DeMorgan’s law} \\ \iff &(p \wedge q \wedge \neg r) \vee \neq ((p \wedge q \wedge \neg r)) && \text{DeMorgan’s law} \\ \iff &T && \text{domination law} \end{align} $$

I had the last step equivalent to the idempotent law $(p \vee p \iff p)$ with: $$ (p \wedge q \wedge \neg r) \vee (p \wedge q \wedge \neg r) $$ which of course doesn't lead me to true or false.

Why is there a $\neq$ sign and why is it true by domination law?

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    $\begingroup$ The sign "$\ne$" is clearly a typo : $(¬p ∨ ¬q ∨ r)$ is equivalent to $\lnot(p ∧ q ∧ ¬r)$. $\endgroup$ – Mauro ALLEGRANZA Oct 14 '14 at 16:43
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    $\begingroup$ If so, $(p∧q∧¬r) \lor ¬(p∧q∧¬r)$ is like $A \lor \lnot A \equiv T$. $\endgroup$ – Mauro ALLEGRANZA Oct 14 '14 at 16:44
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In your second use of DeMorgan's, you should get $\lnot(p \land q \land \lnot r)$ so yu have a tautology of the form $a \lor \lnot a$. Specifically, after your second application of DeMorgan's, you have

$$(p \land q \land \lnot r) \lor \lnot (p \land q \land \lnot r)$$

which is necessarily true, regardless of the truth values of $p, q, r$. Either the left hand side is true, or else its negation (the right-hand side) is true. And since one must be true, the disjunction is thereby true.

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