Question 479319
<pre>
p    q    ~p    ~q    q&#8896;~p 
T    T     F     F     ? 
T    F     F     T     ?
F    T     T     F     ?
F    F     T     T     ?

You need to learn the rules for the four basic 
symbols &#8896;, &#8897;, ->, and <->

Rules:

&#8896; is T only for T&#8896;T, otherwise it's F.

&#8897; is F only for F&#8897;F, otherwise it's T.

-> is F only for T->F, otherwise it's T.

<-> is T for T<->T, F<->F. <-> is F for T<->F, F<->T.

You only have &#8896; here to do, but you need to learn 
those other three rules as well for other truth tables.

Here goes:

p    q    ~p    ~q    q&#8896;~p 
T    T     F     F     ?      
T    F     F     T     ?
F    T     T     F     ?
F    F     T     T     ?

Let's look at q&#8896;~p. 

Let's do the top line:

On the left of the &#8896; in q&#8896;~p is q, and on the top line q has T under it.
On the right of the &#8896; in q&#8896;~p is ~p, and on the top line ~p has F under it.
So on the top line  q&#8896;~p is a case of T&#8896;F. So we look at the rule 
for &#8896;, which is:
"&#8896; is T only for T&#8896;T, otherwise it's F."
T&#8896;F is not a case of T&#8896;T, and therefore it's F.  So we put an F
on the top row underneath q&#8896;~p.  So we now have this:

p    q    ~p    ~q    q&#8896;~p 
T    T     F     F     F      
T    F     F     T     ?
F    T     T     F     ?
F    F     T     T     ?

--------------------------------------------------------------  

Next let's do the second line:

On the left of the &#8896; in q&#8896;~p is q, and on the 2nd line q has F under it.
On the right of the &#8896; in q&#8896;~p is ~p, and on the 2nd line ~p has F under it.
So on the 2nd line  q&#8896;~p is a case of F&#8896;F. So we look at the rule 
for &#8896;, which is:
"&#8896; is T only for T&#8896;T, otherwise it's F."
F&#8896;F is not a case of T&#8896;T, and therefore it's F.  So we put an F
on the 2nd line underneath q&#8896;~p.  So we now have this:

p    q    ~p    ~q    q&#8896;~p 
T    T     F     F     F      
T    F     F     T     F
F    T     T     F     ?
F    F     T     T     ?

--------------------------------------------------------------
 
Next let's do the third line:

On the left of the &#8896; in q&#8896;~p is q, and on the 3rd line q has T under it.
On the right of the &#8896; in q&#8896;~p is ~p, and on the 3rd line ~p has T under it.
So on the 3rd line  q&#8896;~p is a case of T&#8896;T. So we look at the rule 
for &#8896;, which is:
"&#8896; is T only for T&#8896;T, otherwise it's F."
T&#8896;T is INDEED a case of T&#8896;T, and therefore it's T.  So we put a T
on the 3rd row underneath q&#8896;~p.  So we now have this:

p    q    ~p    ~q    q&#8896;~p 
T    T     F     F     F      
T    F     F     T     F
F    T     T     F     T
F    F     T     T     ?

--------------------------------------------------------------

Next let's do the bottom line:

On the left of the &#8896; in q&#8896;~p is q, and on the bottom line q has F under it.
On the right of the &#8896; in q&#8896;~p is ~p, and on the bottom line ~p has T under it.
So on the bottom line q&#8896;~p is a case of F&#8896;T. So we look at the rule 
for &#8896;, which is:
"&#8896; is T only for T&#8896;T, otherwise it's F."
F&#8896;T is not a case of T&#8896;T, and therefore it's F.  So we put an F
on the 2nd row underneath q&#8896;~p.  So finally we have this:

p    q    ~p    ~q    q&#8896;~p 
T    T     F     F     F      
T    F     F     T     F
F    T     T     F     T
F    F     T     T     F

Do you now see how truth tables are done?  You have to learn the
rules for all four symbols &#8896;, &#8897;, ->, and <-> to do truth tables.

Edwin</pre>