3

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

Graham Kemp
  • 133,738
Kagamine Len
  • 526

2 Answers2

1

$\color{green}{\checkmark}$ 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.

$$\begin{align} &\quad (P\wedge (P\implies Q)) \implies Q \\ & \equiv \neg(P\wedge (\neg P\vee Q)) \vee Q & \text{Implication Equivalence} \\ & \equiv (\neg P \vee (P\wedge \neg Q)) \vee Q & \text{DeMorgan's Negation Laws} \\ & \equiv (\neg P \vee \neg Q)\vee Q & \text{Distribution and Identity Laws} \\ & \equiv \neg P\vee (\neg Q\vee Q) & \text{Associativity} \\ & \equiv \top \end{align}$$

Graham Kemp
  • 133,738
0

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

User
  • 295