/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

Problem:
 a(b(x1)) -> b(d(x1))
 a(c(x1)) -> d(d(d(x1)))
 b(d(x1)) -> a(c(b(x1)))
 c(f(x1)) -> d(d(c(x1)))
 d(d(x1)) -> f(x1)
 f(f(x1)) -> a(x1)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [f](x0) = x0 + 2,
   
   [a](x0) = x0 + 3,
   
   [c](x0) = x0,
   
   [b](x0) = 3x0 + 8,
   
   [d](x0) = x0 + 1
  orientation:
   a(b(x1)) = 3x1 + 11 >= 3x1 + 11 = b(d(x1))
   
   a(c(x1)) = x1 + 3 >= x1 + 3 = d(d(d(x1)))
   
   b(d(x1)) = 3x1 + 11 >= 3x1 + 11 = a(c(b(x1)))
   
   c(f(x1)) = x1 + 2 >= x1 + 2 = d(d(c(x1)))
   
   d(d(x1)) = x1 + 2 >= x1 + 2 = f(x1)
   
   f(f(x1)) = x1 + 4 >= x1 + 3 = a(x1)
  problem:
   a(b(x1)) -> b(d(x1))
   a(c(x1)) -> d(d(d(x1)))
   b(d(x1)) -> a(c(b(x1)))
   c(f(x1)) -> d(d(c(x1)))
   d(d(x1)) -> f(x1)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [f](x0) = [0 0 0]x0
              [0 0 1]  ,
    
              [1 1 1]     [0]
    [a](x0) = [0 0 0]x0 + [1]
              [0 0 1]     [0],
    
              [1 0 1]  
    [c](x0) = [0 0 1]x0
              [0 0 1]  ,
    
              [1 0 0]     [1]
    [b](x0) = [0 1 0]x0 + [1]
              [0 0 0]     [0],
    
              [1 0 0]  
    [d](x0) = [0 0 0]x0
              [0 0 1]  
   orientation:
               [1 1 0]     [2]    [1 0 0]     [1]           
    a(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(d(x1))
               [0 0 0]     [0]    [0 0 0]     [0]           
    
               [1 0 3]     [0]    [1 0 0]                
    a(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 = d(d(d(x1)))
               [0 0 1]     [0]    [0 0 1]                
    
               [1 0 0]     [1]    [1 0 0]     [1]              
    b(d(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(c(b(x1)))
               [0 0 0]     [0]    [0 0 0]     [0]              
    
               [1 0 1]      [1 0 1]                
    c(f(x1)) = [0 0 1]x1 >= [0 0 0]x1 = d(d(c(x1)))
               [0 0 1]      [0 0 1]                
    
               [1 0 0]      [1 0 0]          
    d(d(x1)) = [0 0 0]x1 >= [0 0 0]x1 = f(x1)
               [0 0 1]      [0 0 1]          
   problem:
    a(c(x1)) -> d(d(d(x1)))
    b(d(x1)) -> a(c(b(x1)))
    c(f(x1)) -> d(d(c(x1)))
    d(d(x1)) -> f(x1)
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(b) = 4
     w(f) = w(c) = w(d) = w(a) = 0
    status function:
     st(f) = st(c) = st(d) = st(a) = st(b) = [0]
    precedence:
     b > c > d > f ~ a
    problem:
     
    Qed