# Truth table for $(p\land q)\rightarrow r$

I want to know if I'm on the right track with my truth table for (p AND q) implies r....

p | q | r | p AND q | (p AND q) --> r
T   T   T      T        T
T   T   F      T        F
T   F   T      F        T
T   F   F      F        F
F   T   T      F        T
F   T   F      F        F
F   F   T      F        T
F   F   F      F        F


I don't really understand the implication rule when used like this so if anyone can give me a further breakdown that would be great

Thanks!

Let us consider the $r$ and the $(p\land q)$

Just remember the rules for implication

If we have $P\to Q$ (these are just dummy variables to show you the relationship), the possible values are as follows:

The way you wrote your table is correct.

In the first row we have r=T, p AND q = T, and as we know $T\to T = T$, hence you get T.

If we consider the second row, we have r=F, p AND q = T. But remember we are doing $(p\land Q)\to r$, so we have $T\to F$, which is $F$ by the table, and thats how you get F in row 2. Just repeat this for all the rows.

• Ok, So i applied that logic to my table....So the last column should be the following?? T F T T T T T F – Pullingmyhairout May 9 '17 at 0:49

$P \rightarrow Q$ is always True, except when $P$ is True and $Q$ is False.

So, in your case, whenever $p \land q$ is false (rows 3 through 8), $(p \land q) \rightarrow r$ is True