What is the modus ponens of a tautology? 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
 A: $\color{green}{\checkmark}$  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.

$$\begin{align}
&\quad (P\wedge (P\implies Q)) \implies Q 
\\ & \equiv 
\neg(P\wedge (\neg P\vee Q)) \vee Q
& \text{Implication Equivalence}
\\ & \equiv
(\neg P \vee (P\wedge \neg Q)) \vee Q
& \text{DeMorgan's Negation Laws}
\\ & \equiv
(\neg P \vee \neg Q)\vee Q
& \text{Distribution and Identity Laws}
\\ & \equiv
\neg P\vee (\neg Q\vee Q)
& \text{Associativity}
\\ & \equiv
\top
\end{align}$$
A: Everything is correct.
$P \land (P \implies Q)$ is correct.
It can be rewritten as $P \land (\lnot P \lor Q).$
This is exactly the same as just $P \land Q$ because:
As $P$ must be true for the entire proposition to be true, $\lnot P$ cannot be true, i.e $Q$ must be true.
EDIT:
$((P \land (P \implies Q) \implies Q)$ can be rewritten as:
$Q \lor \lnot (P \land ( P \implies Q))$
$Q \lor \lnot(P \land (Q \lor \lnot P))$
$Q \lor \lnot(P \land Q)$
$Q \lor \lnot(\lnot(\lnot P \lor \lnot Q))$
$Q \lor (\lnot P \lor \lnot Q)$
Without parentheses:
$Q \lor \lnot Q \lor \lnot P$
$true \lor \lnot P$
$true$ i.e we have a tautology
