/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:
 strict:
  a(c(c(x1))) -> b(a(a(x1)))
  b(a(b(x1))) -> b(c(a(x1)))
 weak:
  a(a(c(x1))) -> c(a(b(x1)))
  c(c(b(x1))) -> a(b(a(x1)))
  b(c(b(x1))) -> b(a(c(x1)))
  b(c(c(x1))) -> c(b(b(x1)))
  a(a(b(x1))) -> a(c(a(x1)))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 1]  
   [a](x0) = [1 0]x0,
   
             [1 1]     [1]
   [c](x0) = [1 0]x0 + [0],
   
             [1 1]     [0]
   [b](x0) = [1 0]x0 + [1]
  orientation:
                 [3 2]     [3]    [3 2]     [0]              
   a(c(c(x1))) = [2 1]x1 + [2] >= [2 1]x1 + [1] = b(a(a(x1)))
   
                 [3 2]     [1]    [3 2]     [1]              
   b(a(b(x1))) = [2 1]x1 + [2] >= [2 1]x1 + [2] = b(c(a(x1)))
   
                 [3 2]     [2]    [3 2]     [2]              
   a(a(c(x1))) = [2 1]x1 + [1] >= [2 1]x1 + [1] = c(a(b(x1)))
   
                 [3 2]     [3]    [3 2]     [1]              
   c(c(b(x1))) = [2 1]x1 + [2] >= [2 1]x1 + [0] = a(b(a(x1)))
   
                 [3 2]     [2]    [3 2]     [2]              
   b(c(b(x1))) = [2 1]x1 + [3] >= [2 1]x1 + [2] = b(a(c(x1)))
   
                 [3 2]     [3]    [3 2]     [3]              
   b(c(c(x1))) = [2 1]x1 + [3] >= [2 1]x1 + [1] = c(b(b(x1)))
   
                 [3 2]     [1]    [3 2]     [1]              
   a(a(b(x1))) = [2 1]x1 + [1] >= [2 1]x1 + [1] = a(c(a(x1)))
  problem:
   strict:
    b(a(b(x1))) -> b(c(a(x1)))
   weak:
    a(a(c(x1))) -> c(a(b(x1)))
    b(c(b(x1))) -> b(a(c(x1)))
    b(c(c(x1))) -> c(b(b(x1)))
    a(a(b(x1))) -> a(c(a(x1)))
  String Reversal Processor:
   strict:
    b(a(b(x1))) -> a(c(b(x1)))
   weak:
    c(a(a(x1))) -> b(a(c(x1)))
    b(c(b(x1))) -> c(a(b(x1)))
    c(c(b(x1))) -> b(b(c(x1)))
    b(a(a(x1))) -> a(c(a(x1)))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [a](x0) = 2x0 + 1,
     
     [c](x0) = 2x0,
     
     [b](x0) = 2x0 + 2
    orientation:
     b(a(b(x1))) = 8x1 + 12 >= 8x1 + 9 = a(c(b(x1)))
     
     c(a(a(x1))) = 8x1 + 6 >= 8x1 + 4 = b(a(c(x1)))
     
     b(c(b(x1))) = 8x1 + 10 >= 8x1 + 10 = c(a(b(x1)))
     
     c(c(b(x1))) = 8x1 + 8 >= 8x1 + 6 = b(b(c(x1)))
     
     b(a(a(x1))) = 8x1 + 8 >= 8x1 + 5 = a(c(a(x1)))
    problem:
     strict:
      
     weak:
      b(c(b(x1))) -> c(a(b(x1)))
    Qed