In the statement $P$ and $Q$, please show that $\; (P \land (P \Rightarrow Q))\Rightarrow Q \;$ is a tauntology. The state the $\;(P \land (P \Rightarrow Q))\Rightarrow Q\;$ in words.
I know I need to make one of those logic tables. My question is if that the tauntology is all true, I just need to say everything is true right?
So for example:
$$
\begin{array}{}
P & \text{Q} & \text{(P $\Rightarrow$ Q)} & \text{(P $\land$ (P $\Rightarrow$ Q))} & \text{((P $\land$ (P $\Rightarrow$Q)$\Rightarrow$Q} \\
\hline
T & T & T & T & T\\
T & F & F & F & T\\
F & T & T & F & T\\
F & F & T & F & T\\
\end{array}
$$
Like the table I can do but for some odd reason, I can't seem to write it. The definition of a tautology is that all the table result is true so I got that part. But logically I don't follow. I see taht the ending is all true but is the row with $P \land (P \Rightarrow Q)$ correct?
Also when I make the sentence I use the $P$s and the $Q$s right?
Thank you for the help.
(Thank you everybody)
ありがとう 皆さん
Best Answer
$\color{green}{\checkmark}$ All rows are correct.
$P\implies Q$ is true when $Q$ is true or $P$ is false.
$P\wedge (P\implies Q)$ is true when $P$ and $Q$ are both true.
So $(P\wedge (P\implies Q))\implies Q)$ is true when $Q$ is true or $P$ and $Q$ are not both true. Thus it's true when $Q$ is either true or false.
Which means that $(P\wedge (P\implies Q))\implies Q)$ is always true. It's a tautology.
$$\begin{align} &\quad (P\wedge (P\implies Q)) \implies Q \\ & \equiv \neg(P\wedge (\neg P\vee Q)) \vee Q & \text{Implication Equivalence} \\ & \equiv (\neg P \vee (P\wedge \neg Q)) \vee Q & \text{DeMorgan's Negation Laws} \\ & \equiv (\neg P \vee \neg Q)\vee Q & \text{Distribution and Identity Laws} \\ & \equiv \neg P\vee (\neg Q\vee Q) & \text{Associativity} \\ & \equiv \top \end{align}$$