Answer

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<title>Answer</title>
<h1>Answer</h1>

<pre style="font-family:monospace;font-size:large">
  P Q R S  |  (((P&Q)&~S) & (~Q&harr;R))  ~(~R&rarr;S) ((P&rarr;Q)&rarr;(S&rarr;R))
  ---------+-------------------------------------------------
  T T T T  |      T  FF   F  F F     F F T      T  T  T    
  T T T F  |      T  TT   F  F F     F F T      T  T  T    
  T T F T  |      T  FF   F  F T     F T T      T  F  F    
  T T F F  |      T  TT   <b>T</b>  F T     <b>T</b> T F      T  <b>T</b>  T    
  T F T T  |      F  FF   F  T T     F F T      F  T  T    
  T F T F  |      F  FT   F  T T     F F T      F  T  T    
  T F F T  |      F  FF   F  T F     F T T      F  T  F    
  T F F F  |      F  FT   F  T F     T T F      F  T  T    
  F T T T  |      F  FF   F  F F     F F T      T  T  T    
  F T T F  |      F  FT   F  F F     F F T      T  T  T    
  F T F T  |      F  FF   F  F T     F T T      T  F  F    
  F T F F  |      F  FT   F  F T     T T F      T  T  T    
  F F T T  |      F  FF   F  T T     F F T      T  T  T    
  F F T F  |      F  FT   F  T T     F F T      T  T  T    
  F F F T  |      F  FF   F  T F     F T T      T  F  F    
  F F F F  |      F  FT   F  T F     T T F      T  T  T    

</pre>

This sequent is valid. Notice that the first premise is true only under the 4th assignment. This means that no other assignment can be an invalidating assignment, because an invalidating assignment must be one where <b>all</b> the premises are true.
<p> So in this particular example, once we have found out that the first premise is true only under the 4th assignment, we only need to calculate the truth-values of the second premise and the conclusion under the same assignment. There is no need to draw the rest of the truth-table. This means we can determine validity a lot faster.

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