1. ## Boolean Expression

How can i Make truth table for the following expression
~p→~r V q ~ p V r

great confusion to start because there is no brackets

2. ## Re: Boolean Expression

Originally Posted by suhail
How can i Make truth table for the following expression
~p→~r V q ~ p V r

great confusion to start because there is no brackets
I'm no logic expert but I believe this can be rewritten as

$(\neg p \rightarrow \neg r) \vee \left(q \wedge (p \vee r)\right)$

$\neg p$ is the same as $\sim p$ i.e. NOT p

3. ## Re: Boolean Expression

I confirm that $\land$ usually binds stronger than $\lor$. Negation usually binds the stronest, i.e., it pertains only to the smallest following subformula. With respect to $\to$ and $\lor$, conventions may vary. I would parse $A\to B\lor C$ as $A\to (B\lor C)$. In any case, precise rules for parsing formulas should be described in the textbook or lecture notes.

4. ## Re: Boolean Expression

Originally Posted by romsek
I'm no logic expert but I believe this can be rewritten as

$(\neg p \rightarrow \neg r) \vee \left(q \wedge (p \vee r)\right)$

$\neg p$ is the same as $\sim p$ i.e. NOT p
Originally Posted by emakarov
I confirm that $\land$ usually binds stronger than $\lor$. Negation usually binds the stronest, i.e., it pertains only to the smallest following subformula. With respect to $\to$ and $\lor$, conventions may vary. I would parse $A\to B\lor C$ as $A\to (B\lor C)$. In any case, precise rules for parsing formulas should be described in the textbook or lecture notes.
Yes, you are right. It should be

$\left(\neg p \rightarrow (\neg r \vee q)\right) \wedge (p \vee r)$