SOLUTION: I am a little confussed. Construct a truth table for q -> ~p

q->~p

Start with this since the expression involves four things
p, q, ~q, and q->~p

| p | q | ~p | q->~p |
|---|---|----|-------|
|   |   |    |       |
|   |   |    |       |
|   |   |    |       |
|   |   |    |       |

Under p write TTFF, under q list TFTF

| p | q | ~p | q->~p |
|---|---|----|-------|
| T | T |    |       |
| T | F |    |       |
| F | T |    |       |
| F | F |    |       |

The rule for ~↏ 
if there is a T in the box put an F
if there is an F in the box put a T
That is ~T becomes F and ~F becoms T.
Looking at the column under p we follow
that rule and get FFTT:

| p | q | ~p | q->~p |
|---|---|----|-------|
| T | T |  F |       |
| T | F |  F |       |
| F | T |  T |       |
| F | F |  T |       |


The rule for ↏->↏ 
if there is a T in the first box and an F
in the second box, the expression is F.
That is, if you see T->F the expression
becomes F, otherwise the expression is T.
So in any other case, the expression 
is T. Looking at the column under q and ~p we 
follow that rule and get FTTT

| p | q | ~p | q->~p |
|---|---|----|-------|
| T | T |  F |   F   |
| T | F |  F |   T   |
| F | T |  T |   T   |
| F | F |  T |   T   |

Notice that only F is in the first case because
it was the only case where q is T and ~p is F.

Edwin