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


--------------------------------------------------------------------------------


YES

Problem:
 b(a(a(a(x1)))) -> a(a(b(a(x1))))
 b(a(b(b(x1)))) -> b(a(a(a(x1))))
 a(a(b(b(x1)))) -> a(b(b(a(x1))))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [b](x0) = x0 + 1,
   
   [a](x0) = x0
  orientation:
   b(a(a(a(x1)))) = x1 + 1 >= x1 + 1 = a(a(b(a(x1))))
   
   b(a(b(b(x1)))) = x1 + 3 >= x1 + 1 = b(a(a(a(x1))))
   
   a(a(b(b(x1)))) = x1 + 2 >= x1 + 2 = a(b(b(a(x1))))
  problem:
   b(a(a(a(x1)))) -> a(a(b(a(x1))))
   a(a(b(b(x1)))) -> a(b(b(a(x1))))
  String Reversal Processor:
   a(a(a(b(x1)))) -> a(b(a(a(x1))))
   b(b(a(a(x1)))) -> a(b(b(a(x1))))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 1]     [0]
     [b](x0) = [0 0 1]x0 + [0]
               [0 1 0]     [1],
     
               [1 0 0]     [0]
     [a](x0) = [0 1 0]x0 + [1]
               [0 0 0]     [0]
    orientation:
                      [1 0 1]     [0]    [1 0 0]     [0]                 
     a(a(a(b(x1)))) = [0 0 1]x1 + [3] >= [0 0 0]x1 + [1] = a(b(a(a(x1))))
                      [0 0 0]     [0]    [0 0 0]     [0]                 
     
                      [1 1 0]     [3]    [1 1 0]     [2]                 
     b(b(a(a(x1)))) = [0 1 0]x1 + [3] >= [0 1 0]x1 + [3] = a(b(b(a(x1))))
                      [0 0 0]     [1]    [0 0 0]     [0]                 
    problem:
     a(a(a(b(x1)))) -> a(b(a(a(x1))))
    KBO Processor:
     weight function:
      w0 = 1
      w(b) = 1
      w(a) = 0
     precedence:
      a > b
     problem:
      
     Qed