Question 451893
<pre>

Start with this putting TTFF under p and TFTF under q 

p  q  ~p  ~q   &#8764;p&#8744;q  (&#8764;p&#8744;q)&#8594; &#8764;q
—————————————————————————————————
T  T
T  F
F  T
F  F

Under the ~p put the opposite of p. Since p has TTFF,
~p will have FFTT:

p  q  ~p  ~q  &#8764;p&#8744;q  (&#8764;p&#8744;q)&#8594; &#8764;q
————————————————————————————————
T  T   F
T  F   F
F  T   T
F  F   T

Under the ~q put the opposite of q. Since q has TFTF,
~q will have FTFT:

p  q  ~p  ~q  &#8764;p&#8744;q  (&#8764;p&#8744;q)&#8594; &#8764;q
————————————————————————————————
T  T   F   F
T  F   F   T
F  T   T   F 
F  F   T   T

We make &#8764;p&#8744;q out of the columns ~p, q according to
the rule:

If there are F's on both sides of the disjunction
symbol &#8744; the disjunction is F, otherwise it's T
Since ~p is FFTT and q is TFTF, then only the 2nd 
row has F under both ~q and p, so it gets F and the
other three get T, so under &#8764;p&#8744;q we put TFTT

p  q  ~p  ~q  &#8764;p&#8744;q  (&#8764;p&#8744;q)&#8594; &#8764;q
————————————————————————————————
T  T   F   F    T 
T  F   F   T    F
F  T   T   F    T
F  F   T   T    T


We make (&#8764;p&#8744;q)&#8594;&#8764;q out of the columns &#8764;p&#8744;q and &#8764;q according to
the rule:

If there is a T on the left of the conditional 
symbol &#8594; and a T on the right side of the conditional
symbol, the conditional is F, otherwise it's T.
Since &#8764;p&#8744;q is TFTT and ~q is FTFT, then the 1st 
and 3rd rows has F have this, so it gets F in rows
1 and 3
other two get T, so under (&#8764;p&#8744;q)&#8594; &#8764;q
we put FTFT

p  q  ~p  ~q  &#8764;p&#8744;q  (&#8764;p&#8744;q)&#8594; &#8764;q
————————————————————————————————
T  T   F   F    T         F 
T  F   F   T    F         T
F  T   T   F    T         F
F  F   T   T    T         T

Edwin</pre>