/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(b(a(x1))) -> b(a(x1))
 b(b(b(x1))) -> b(a(b(x1)))
 a(a(x1)) -> b(b(b(x1)))

Proof:
 String Reversal Processor:
  a(b(a(x1))) -> a(b(x1))
  b(b(b(x1))) -> b(a(b(x1)))
  a(a(x1)) -> b(b(b(x1)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
              [1 0]  
    [b](x0) = [0 0]x0,
    
              [1 2]     [0]
    [a](x0) = [2 0]x0 + [2]
   orientation:
                  [1 2]     [0]    [1 0]     [0]           
    a(b(a(x1))) = [2 4]x1 + [2] >= [2 0]x1 + [2] = a(b(x1))
    
                  [1 0]      [1 0]                
    b(b(b(x1))) = [0 0]x1 >= [0 0]x1 = b(a(b(x1)))
    
               [5 2]     [4]    [1 0]                
    a(a(x1)) = [2 4]x1 + [2] >= [0 0]x1 = b(b(b(x1)))
   problem:
    a(b(a(x1))) -> a(b(x1))
    b(b(b(x1))) -> b(a(b(x1)))
   KBO Processor:
    weight function:
     w0 = 1
     w(b) = w(a) = 1
    precedence:
     b > a
    problem:
     
    Qed