Is this a correct application of logical Absorption and Reduction?

Question

Absorption:

$$P \land (P \lor Q) = P$$

Reduction:

$$P \land (\neg P \lor Q) = P \land Q$$

Starting with,

$[\neg P \land (\neg P \lor Q)\land (P\lor \neg Q)\lor \neg P \land (\neg R \lor Q)\land (R\lor \neg Q)]\lor [R \land (\neg P \lor Q)\land (P \lor \neg Q)\land R \land (\neg R \lor Q)\land (R \lor \neg )]=$

[Application of Absorption on $\neg P \land (\neg P \lor Q)$, by letting $P= \neg P, Q=Q$ and by commutation on $(R \lor \neg Q)].$

$[\neg P \land (P \lor \neg ) \lor \neg P \land (\neg R \lor Q)\land (R \lor \neg Q)]\lor [R \land(\neg P \lor Q)\land (P \lor \neg Q)\land R \land (R \lor \neg Q)\land (\neg R \lor Q)]=$

[Application of Absorption on $R \land (R \lor \neg Q$), by letting $P=R, Q=\neg Q$]

$[\neg P \land (P \lor \neg Q)\lor \neg P \land (\neg R \lor Q)\land (R \lor \neg Q)]\lor[R \land (\neg P \lor Q) \land (P \lor \neg Q)]\land R \land (\neg R \lor Q)]=$

[Using the rule of Reduction on $\neg P \land (P \lor \neg Q)$, by letting $P = \neg P,Q= \neg Q]$

$[\neg P \land \neg Q)\lor \neg P \land (\neg R \lor Q)\land (R \lor \neg Q)]\lor [R \land (\neg P \lor Q)\land (P \lor \neg Q)\land R \land(\neg R \lor Q)]=$

[Using the rule of Reduction on $R \land (\neg R \lor Q)$, by letting P=R,Q=Q]

$[(\neg P \land \neg Q) \lor \neg P \land (\neg R \lor Q) \land (R \lor \neg Q)]\lor [R \land (\neg P \lor Q)\land (P \lor \neg Q)\land (R \land Q)]$

Answer 1

First, you need some parentheses, since you have one of those $X \land Y \lor Z$ constructions. So:

$[\color{red}(\neg P \land (\neg P \lor Q)\land (P\lor \neg Q)\color{red})\lor \color{red}(\neg P \land (\neg R \lor Q)\land (R\lor \neg Q)\color{red})]\lor [R \land (\neg P \lor Q)\land (P \lor \neg Q)\land R \land (\neg R \lor Q)\land (R \lor \neg Q)]$

Likewise, seeing where this problem comes from, one of the $\land$'s of the right square bracketed term should be a $\lor$, and so you likewise get:

$[(\neg P \land (\neg P \lor Q)\land (P\lor \neg Q))\lor (\neg P \land (\neg R \lor Q)\land (R\lor \neg Q))]\lor [\color{red}(R \land (\neg P \lor Q)\land (P \lor \neg Q)\color{red})\color{red}\lor \color{red}(R \land (\neg R \lor Q)\land (R \lor \neg Q)\color{red})]$

Because you were missing parentheses, I actually intially misread your statements, and concluded that you had applied the rules incorrectly, but as it turns out you did apply them correctly!

Here is what it looks like with proper parentheses in place. First, the $\neg P$ term absorbs the $(\neg P \lor Q)$ term, resulting in:

$[(\neg P \land (P\lor \neg Q))\lor (\neg P \land (\neg R \lor Q)\land (R\lor \neg Q))]\lor [(R \land (\neg P \lor Q)\land (P \lor \neg Q))\lor (R \land (\neg R \lor Q)\land (R \lor \neg Q))]$

Likewise, the $R$ completely gets rid of the $(R \lor \neg Q)$:

$[(\neg P \land (P\lor \neg Q))\lor (\neg P \land (\neg R \lor Q)\land (R\lor \neg Q))]\lor [(R \land (\neg P \lor Q)\land (P \lor \neg Q))\lor (R \land (\neg R \lor Q))]$

Now, Reduction using the $\neg P$:

$[(\neg P \land \neg Q)\lor (\neg P \land (\neg R \lor Q)\land (R\lor \neg Q))]\lor [(R \land (\neg P \lor Q)\land (P \lor \neg Q))\lor (R \land (\neg R \lor Q))]$

And finally, Reduction using the $R$:

$[(\neg P \land \neg Q)\lor (\neg P \land (\neg R \lor Q)\land (R\lor \neg Q))]\lor [(R \land (\neg P \lor Q)\land (P \lor \neg Q))\lor (R \land Q))]$

Good job! ... just make sure parentheses fully disambiguate everything