0
$\begingroup$

Show that the following logical expression is universally valid.

$$ (p \lor (q \land r)) \land p \iff ( \neg p \lor (q \land r) \implies p) $$

Here's what I tried so far:

$$[ \ (p \lor (q \land r)) \land p \iff ( \neg p \lor (q \land r) \implies p) \ ]^{\displaystyle \beta } = T$$

Using the definition of $ \iff$, I get

$$ [ \ (p \lor (q \land r)) \land p \ ]^{\displaystyle \beta } = \ [ \ ( \neg p \lor (q \land r) \implies p) \ ]^{\displaystyle \beta } = T $$

Now I would have to distinguish between two cases, namely when $\displaystyle \beta(p) = T$ and when $\displaystyle \beta(p) = F$ and here is my problem. When I start to show the first case, I end up getting $[ \ (p \lor (q \land r)) \land p \ ]^{\displaystyle \beta } = [ \ p \lor (q \land r) \ ]^{\displaystyle \beta }$ but I don't know how I got rid of the $ \ \land p \ $ operation.

$\endgroup$
4
  • $\begingroup$ Are you allowed in this to use truth tables? $\endgroup$
    – drhab
    Commented Dec 12, 2018 at 12:17
  • $\begingroup$ Unfortunately no. We have to specifically use the variable assignment method. $\endgroup$
    – user496674
    Commented Dec 12, 2018 at 12:18
  • $\begingroup$ Are you sure you copied the question correctly? How many $\lnot$ in the original? $\endgroup$
    – DanielV
    Commented Dec 12, 2018 at 13:04
  • $\begingroup$ Can you show that both sides of the $\iff$ are equivalent to just $p$? $\endgroup$ Commented Dec 12, 2018 at 17:34

2 Answers 2

0
$\begingroup$

We have $$ \begin{array}{|c|c|c| c|c|c| c|} \hline p & q & r & q\land r & p \lor (q\land r) & (p \lor (q\land r))\land p \\\hline T & T & T & T & T & T \\\hline T & T & F & F & T & T \\\hline T & F & T & T & T & T \\\hline T & F & F & F & T & T \\\hline F & T & T & T & T & F \\\hline F & T & F & F & F & F \\\hline F & F & T & F & F & F \\\hline F & F & F & F & F & F \\\hline \end{array} $$ and $$ \begin{array}{|c|c|c| c|c|c|c| c|} \hline p & q & r & \lnot p & (q\land r) & ( \neg p \lor (q \land r)) & ( \neg p \lor (q \land r)) \implies p \\\hline T & T & T & F & T & T & T \\\hline T & T & F & F & F & F & T \\\hline T & F & T & F & F & F & T \\\hline T & F & F & F & F & F & T \\\hline F & T & T & T & T & T & F \\\hline F & T & F & T & F & T & F \\\hline F & F & T & T & F & T & F \\\hline F & F & F & T & F & T & F \\\hline \end{array} $$

$\endgroup$
0
$\begingroup$

Hint

If you do not want to use truth table, you can reason by contradiction.

Assume that :

$(p \lor (q \land r)) \land p \iff ( \neg p \lor (q \land r) \implies p)$

is not valid.

This means that, for some variable assignment $\beta$ we have :

$[(p \lor (q \land r)) \land p]^{\beta} = \text T \text { and } [(\neg p \lor (q \land r) \implies p)]^{\beta} = \text F$.

$\endgroup$

You must log in to answer this question.