Question 680763
<pre>
We'll do each case separately then we'll
put the four cases into a truth table:

(P &#8594; Q) &#8596; (星 &#8744; Q)

Let "T" mean "true" and let "F" be "false:

Case 1:  P is true and Q is true:

(P &#8594; Q) &#8596; (星 &#8744; Q)
(T &#8594; T) &#8596; (曷 &#8744; T)
   T    &#8596; ( F &#8744; T)
   T    &#8596;     T
        T

Case 2:  P is true and Q is false:

(P &#8594; Q) &#8596; (星 &#8744; Q)
(T &#8594; F) &#8596; (曷 &#8744; F)
   F    &#8596; ( F &#8744; F)
   F    &#8596;     F
        T

Case 3:  P is false and Q is true:

(P &#8594; Q) &#8596; (星 &#8744; Q)
(F &#8594; T) &#8596; (政 &#8744; T)
   T    &#8596; ( T &#8744; T)
   T    &#8596;     T
        T  

Case 4:  P is false and Q is false:

(P &#8594; Q) &#8596; (星 &#8744; Q)
(F &#8594; F) &#8596; (政 &#8744; F)
   T    &#8596; ( T &#8744; F)
   T    &#8596;     T
        T  

It is true in all four cases, so it is a tautology.

Or you can do it as a truth table which is equivalent to
the four cases above:

        <u>| P | Q | 星 | P &#8594; Q | 星 &#8744; Q | (P &#8594; Q) &#8596; (星 &#8744; Q |</u>
case 1: | T | T |  F |   T   |    T   |         T          | 
case 2: | T | F |  F |   F   |    F   |         T          |  
case 3: | F | T |  T |   T   |    T   |         T          |  
case 4: | F | F |  T |   T   |    T   |         T          |    

Since the last column came out all trues, the original statement
is a tautology.

Edwin</pre>