Can this logic expression be further simplified? I am learning logic at faculty and of course it causes a little bit of confusion sometimes.    
$P \lor \neg Q \lor (P \land \neg R)$
Am I forgetting a law (probably yes)?
 A: The solution in the comments is a smart human solution, which could be very difficult to find for a big general logical expression. But there is an algorithm that almost always works (to a point) but that sometimes is tedious:


*

*Express all logical connectives expressed in XOR, AND and TRUE $: (+,\cdot,1)$

*Use the algebraic laws (commutativity, associativity, distributivity, additive and multiplicative units, idempotence $X\cdot X=X$, and the additive inverse $X+X=0$) in Boolean rings to simplify almost like for numbers.

*Eventually try to simplify by substitute back to other connectives as below.


\begin{array}{l|l}
connective & substitute \\
\hline
\neg X& 1+X\\
X\wedge Y& XY\\
X\vee Y& X+Y+XY\\
X\oplus Y&X+Y\\
X\Rightarrow Y&1+X+XY\\
X\Leftrightarrow Y&1+X+Y
\end{array}
Your example (tedious):
$(P\vee\neg Q)\vee(P\wedge\neg R)\equiv(P+(1+Q)+P(1+Q))\vee(P(1+R))\equiv$
$(P+1+Q+P+PQ)\vee(P+PR)\equiv(1+Q+PQ)\vee(P+PR)\equiv$
$(1+Q+PQ)+(P+PR)+(1+Q+PQ)(P+PR)\equiv\\1+Q+PQ+P+PR+P+PR+QP+QPR+PQP+PQPR\equiv$
$1+P+P+Q+PQ+PQ+PQ+PR+PR+QPR+QPR\equiv\\1+Q+PQ\equiv(\neg Q\vee P)\equiv(Q\Rightarrow P)$
