/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:
 a(c(a(x1))) -> c(a(c(x1)))
 a(a(b(x1))) -> a(d(b(x1)))
 a(b(x1)) -> b(a(a(x1)))
 d(d(x1)) -> a(d(b(x1)))
 b(b(x1)) -> b(c(x1))
 a(d(c(x1))) -> c(a(x1))
 b(c(x1)) -> a(a(a(x1)))

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