/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

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

Proof:
 String Reversal Processor:
  a(a(x1)) -> c(b(x1))
  b(b(x1)) -> d(c(x1))
  b(x1) -> a(x1)
  c(c(x1)) -> f(d(x1))
  d(d(x1)) -> f(f(f(x1)))
  d(x1) -> b(x1)
  f(f(x1)) -> a(g(x1))
  g(g(x1)) -> a(x1)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 1 0]     [0]
    [g](x0) = [0 0 1]x0 + [1]
              [0 0 0]     [0],
    
              [1 1 0]  
    [f](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 1]  
    [d](x0) = [0 0 0]x0
              [0 1 0]  ,
    
              [1 0 1]  
    [b](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 1]  
    [c](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 1]  
    [a](x0) = [0 0 0]x0
              [0 0 0]  
   orientation:
               [1 0 1]      [1 0 1]             
    a(a(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(b(x1))
               [0 0 0]      [0 0 0]             
    
               [1 0 1]      [1 0 1]             
    b(b(x1)) = [0 0 0]x1 >= [0 0 0]x1 = d(c(x1))
               [0 0 0]      [0 0 0]             
    
            [1 0 1]      [1 0 1]          
    b(x1) = [0 0 0]x1 >= [0 0 0]x1 = a(x1)
            [0 0 0]      [0 0 0]          
    
               [1 0 1]      [1 0 1]             
    c(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = f(d(x1))
               [0 0 0]      [0 0 0]             
    
               [1 1 1]      [1 1 0]                
    d(d(x1)) = [0 0 0]x1 >= [0 0 0]x1 = f(f(f(x1)))
               [0 0 0]      [0 0 0]                
    
            [1 0 1]      [1 0 1]          
    d(x1) = [0 0 0]x1 >= [0 0 0]x1 = b(x1)
            [0 1 0]      [0 0 0]          
    
               [1 1 0]      [1 1 0]             
    f(f(x1)) = [0 0 0]x1 >= [0 0 0]x1 = a(g(x1))
               [0 0 0]      [0 0 0]             
    
               [1 1 1]     [1]    [1 0 1]          
    g(g(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 = a(x1)
               [0 0 0]     [0]    [0 0 0]          
   problem:
    a(a(x1)) -> c(b(x1))
    b(b(x1)) -> d(c(x1))
    b(x1) -> a(x1)
    c(c(x1)) -> f(d(x1))
    d(d(x1)) -> f(f(f(x1)))
    d(x1) -> b(x1)
    f(f(x1)) -> a(g(x1))
   String Reversal Processor:
    a(a(x1)) -> b(c(x1))
    b(b(x1)) -> c(d(x1))
    b(x1) -> a(x1)
    c(c(x1)) -> d(f(x1))
    d(d(x1)) -> f(f(f(x1)))
    d(x1) -> b(x1)
    f(f(x1)) -> g(a(x1))
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [g](x0) = x0,
      
      [f](x0) = x0 + 4,
      
      [d](x0) = x0 + 8,
      
      [b](x0) = x0 + 8,
      
      [c](x0) = x0 + 8,
      
      [a](x0) = x0 + 8
     orientation:
      a(a(x1)) = x1 + 16 >= x1 + 16 = b(c(x1))
      
      b(b(x1)) = x1 + 16 >= x1 + 16 = c(d(x1))
      
      b(x1) = x1 + 8 >= x1 + 8 = a(x1)
      
      c(c(x1)) = x1 + 16 >= x1 + 12 = d(f(x1))
      
      d(d(x1)) = x1 + 16 >= x1 + 12 = f(f(f(x1)))
      
      d(x1) = x1 + 8 >= x1 + 8 = b(x1)
      
      f(f(x1)) = x1 + 8 >= x1 + 8 = g(a(x1))
     problem:
      a(a(x1)) -> b(c(x1))
      b(b(x1)) -> c(d(x1))
      b(x1) -> a(x1)
      d(x1) -> b(x1)
      f(f(x1)) -> g(a(x1))
     KBO Processor:
      weight function:
       w0 = 1
       w(g) = w(f) = w(d) = w(b) = w(a) = 1
       w(c) = 0
      precedence:
       c > d > f ~ b > g ~ a
      problem:
       
      Qed