# Show that $(p \lor (q \land r)) \land p \iff ( \neg p \lor (q \land r) \implies p)$ is valid.

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.

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

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

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 :

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