/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:
 1(2(1(x1))) -> 2(0(2(x1)))
 0(2(1(x1))) -> 1(0(2(x1)))
 L(2(1(x1))) -> L(1(0(2(x1))))
 1(2(0(x1))) -> 2(0(1(x1)))
 1(2(R(x1))) -> 2(0(1(R(x1))))
 0(2(0(x1))) -> 1(0(1(x1)))
 L(2(0(x1))) -> L(1(0(1(x1))))
 0(2(R(x1))) -> 1(0(1(R(x1))))

Proof:
 String Reversal Processor:
  1(2(1(x1))) -> 2(0(2(x1)))
  1(2(0(x1))) -> 2(0(1(x1)))
  1(2(L(x1))) -> 2(0(1(L(x1))))
  0(2(1(x1))) -> 1(0(2(x1)))
  R(2(1(x1))) -> R(1(0(2(x1))))
  0(2(0(x1))) -> 1(0(1(x1)))
  0(2(L(x1))) -> 1(0(1(L(x1))))
  R(2(0(x1))) -> R(1(0(1(x1))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 1]  
    [R](x0) = [0 0 1]x0
              [1 0 0]  ,
    
              [1 0 0]  
    [L](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 0]  
    [0](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 0]     [0]
    [2](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [1],
    
              [1 0 0]  
    [1](x0) = [0 0 0]x0
              [0 0 1]  
   orientation:
                  [1 0 0]     [0]    [1 0 0]     [0]              
    1(2(1(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 2(0(2(x1)))
                  [0 0 0]     [1]    [0 0 0]     [1]              
    
                  [1 0 0]     [0]    [1 0 0]     [0]              
    1(2(0(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 2(0(1(x1)))
                  [0 0 0]     [1]    [0 0 0]     [1]              
    
                  [1 0 0]     [0]    [1 0 0]     [0]                 
    1(2(L(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 2(0(1(L(x1))))
                  [0 0 0]     [1]    [0 0 0]     [1]                 
    
                  [1 0 0]      [1 0 0]                
    0(2(1(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(2(x1)))
                  [0 0 0]      [0 0 0]                
    
                  [1 0 0]     [1]    [1 0 0]                   
    R(2(1(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = R(1(0(2(x1))))
                  [1 0 0]     [0]    [1 0 0]                   
    
                  [1 0 0]      [1 0 0]                
    0(2(0(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(1(x1)))
                  [0 0 0]      [0 0 0]                
    
                  [1 0 0]      [1 0 0]                   
    0(2(L(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(1(L(x1))))
                  [0 0 0]      [0 0 0]                   
    
                  [1 0 0]     [1]    [1 0 0]                   
    R(2(0(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = R(1(0(1(x1))))
                  [1 0 0]     [0]    [1 0 0]                   
   problem:
    1(2(1(x1))) -> 2(0(2(x1)))
    1(2(0(x1))) -> 2(0(1(x1)))
    1(2(L(x1))) -> 2(0(1(L(x1))))
    0(2(1(x1))) -> 1(0(2(x1)))
    0(2(0(x1))) -> 1(0(1(x1)))
    0(2(L(x1))) -> 1(0(1(L(x1))))
   String Reversal Processor:
    1(2(1(x1))) -> 2(0(2(x1)))
    0(2(1(x1))) -> 1(0(2(x1)))
    L(2(1(x1))) -> L(1(0(2(x1))))
    1(2(0(x1))) -> 2(0(1(x1)))
    0(2(0(x1))) -> 1(0(1(x1)))
    L(2(0(x1))) -> L(1(0(1(x1))))
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [L](x0) = 2x0 + 8,
      
      [0](x0) = x0 + 2,
      
      [2](x0) = 4x0 + 2,
      
      [1](x0) = 2x0 + 2
     orientation:
      1(2(1(x1))) = 16x1 + 22 >= 16x1 + 18 = 2(0(2(x1)))
      
      0(2(1(x1))) = 8x1 + 12 >= 8x1 + 10 = 1(0(2(x1)))
      
      L(2(1(x1))) = 16x1 + 28 >= 16x1 + 28 = L(1(0(2(x1))))
      
      1(2(0(x1))) = 8x1 + 22 >= 8x1 + 18 = 2(0(1(x1)))
      
      0(2(0(x1))) = 4x1 + 12 >= 4x1 + 10 = 1(0(1(x1)))
      
      L(2(0(x1))) = 8x1 + 28 >= 8x1 + 28 = L(1(0(1(x1))))
     problem:
      L(2(1(x1))) -> L(1(0(2(x1))))
      L(2(0(x1))) -> L(1(0(1(x1))))
     Bounds Processor:
      bound: 1
      enrichment: match
      automaton:
       final states: {5}
       transitions:
        00(5) -> 5*
        L1(14) -> 15*
        11(16) -> 17*
        11(13) -> 14*
        01(12) -> 13*
        21(11) -> 12*
        L0(5) -> 5*
        20(5) -> 5*
        10(5) -> 5*
        5 -> 16,11
        15 -> 5*
        17 -> 12*
      problem:
       
      Qed