Showing that $\lnot Q \lor (\lnot Q \land R) = \lnot Q$ without a truth table I've done a truth table after reducing it to this and it seems to be equal to $\neg Q$:
$$\lnot Q \lor (\lnot Q \land R) = \lnot Q$$
But when I try to show it without a truth table (with just transformations), I end up in a loop between that and:
$$\lnot Q \land (\lnot Q \lor R)$$
Is there a way to show this is true without using a truth table? What am I missing?!
Thanks in advance!
 A: When it comes to showing that logical statements are independent of certain variables I like to cycle through them and show it this way. Notice that you've shown that your statement is independent of the truth or falsity of $R$ so I'll take that approach here.
Let's suppose $R = T$, then
$$\neg Q\vee(\neg Q\wedge R) = \neg Q\vee(\neg Q\wedge T) = \neg Q\vee (\neg Q)= \neg Q.$$
Likewise if $R = F$, then
$$\neg Q\vee(\neg Q\wedge R) = \neg Q\vee(\neg Q\wedge F) = \neg Q\vee F = \neg Q.$$
Thus no matter what the truth value of $R$ is we get $\neg Q$ so we are forced to conclude that $\neg Q\vee(\neg Q\wedge R) = \neg Q.$
A: \begin{align*}
\neg Q \vee (\neg Q\wedge R)
&= (\neg Q\wedge 1) \vee (\neg Q\wedge R) \\
&= \neg Q \wedge (1\vee R) \\
&= \neg Q \wedge 1 \\
&= \neg Q
\end{align*}
A: $Q\implies\lnot Q \wedge R\implies\lnot Q$
Contradiction!
A: Assume that Q is true.  Then it follows that $\lnot$Q is false, and ($\lnot$Q$\land$R) is false also.  So, ¬Q∨(¬Q∧R) is false.
Assume that Q is false.  Then $\lnot$Q is true, and thus so is ¬Q∨(¬Q∧R).
Since Q is either true or false, and since in either case the equality here ¬Q∨(¬Q∧R)=¬Q, it follows that ¬Q∨(¬Q∧R)=¬Q in all cases.
A: If you don't mind using order theory, you can show it without any reference to true or false. Let $\leq$ be an order with $A \vee B \geq A$ and $A \wedge B \leq A$. Then, $\neg Q \vee (\neg Q \wedge R) \geq \neg Q$ and $\neg Q \wedge (\neg Q \wedge R) \leq \neg Q$. If the two expressions are equal, you can see that they must equal $\neg Q$.
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$Here is yet another way to calculate this:
$$\calc
\lnot Q \lor (\lnot Q \land R) \;\equiv\; \lnot Q
\calcop\equiv{both $\;X \lor Y \equiv Y\;$ and $\;\lnot X \lor Y\;$ are alternative ways to write $\;X \Rightarrow Y\;$}
\lnot (\lnot Q \land R) \lor \lnot Q
\calcop\equiv{DeMorgan}
Q \lor \lnot R \lor \lnot Q
\calcop\equiv{excluded middle; simplify}
\true
\endcalc$$
