/export/starexec/sandbox/solver/bin/starexec_run_ttt2 /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

Problem:
 f(a(),a()) -> f(a(),b())
 f(a(),b()) -> f(s(a()),c())
 f(s(X),c()) -> f(X,c())
 f(c(),c()) -> f(a(),a())

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
         [0]
   [c] = [0]
         [1],
   
                 [1 0 1]     [1 1 0]     [0]
   [f](x0, x1) = [0 1 1]x0 + [0 0 0]x1 + [0]
                 [0 0 0]     [0 0 0]     [1],
   
             [1 0 0]  
   [s](x0) = [1 1 0]x0
             [0 0 1]  ,
   
         [0]
   [a] = [1]
         [0],
   
         [0]
   [b] = [0]
         [0]
  orientation:
                [1]    [0]             
   f(a(),a()) = [1] >= [1] = f(a(),b())
                [1]    [1]             
   
                [0]    [0]                
   f(a(),b()) = [1] >= [1] = f(s(a()),c())
                [1]    [1]                
   
                 [1 0 1]    [0]    [1 0 1]    [0]           
   f(s(X),c()) = [1 1 1]X + [0] >= [0 1 1]X + [0] = f(X,c())
                 [0 0 0]    [1]    [0 0 0]    [1]           
   
                [1]    [1]             
   f(c(),c()) = [1] >= [1] = f(a(),a())
                [1]    [1]             
  problem:
   f(a(),b()) -> f(s(a()),c())
   f(s(X),c()) -> f(X,c())
   f(c(),c()) -> f(a(),a())
  Matrix Interpretation Processor: dim=3
   
   interpretation:
          [0]
    [c] = [0]
          [0],
    
                  [1 0 0]     [1 0 0]  
    [f](x0, x1) = [0 0 0]x0 + [1 0 0]x1
                  [0 0 0]     [0 0 0]  ,
    
              [1 0 0]  
    [s](x0) = [0 0 1]x0
              [0 0 0]  ,
    
          [0]
    [a] = [0]
          [1],
    
          [1]
    [b] = [0]
          [0]
   orientation:
                 [1]    [0]                
    f(a(),b()) = [1] >= [0] = f(s(a()),c())
                 [0]    [0]                
    
                  [1 0 0]     [1 0 0]            
    f(s(X),c()) = [0 0 0]X >= [0 0 0]X = f(X,c())
                  [0 0 0]     [0 0 0]            
    
                 [0]    [0]             
    f(c(),c()) = [0] >= [0] = f(a(),a())
                 [0]    [0]             
   problem:
    f(s(X),c()) -> f(X,c())
    f(c(),c()) -> f(a(),a())
   Matrix Interpretation Processor: dim=3
    
    interpretation:
           [0]
     [c] = [0]
           [1],
     
                   [1 0 1]     [1 0 0]  
     [f](x0, x1) = [0 1 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  ,
     
                    [0]
     [s](x0) = x0 + [0]
                    [1],
     
           [0]
     [a] = [0]
           [0]
    orientation:
                   [1 0 1]    [1]    [1 0 1]            
     f(s(X),c()) = [0 1 0]X + [0] >= [0 1 0]X = f(X,c())
                   [0 0 0]    [0]    [0 0 0]            
     
                  [1]    [0]             
     f(c(),c()) = [0] >= [0] = f(a(),a())
                  [0]    [0]             
    problem:
     
    Qed