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

Proof:
 String Reversal Processor:
  a(b(x1)) -> b(a(x1))
  a(a(a(x1))) -> b(a(a(b(x1))))
  b(b(b(b(x1)))) -> a(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [b](x0) = x0 + 4,
    
    [a](x0) = x0 + 8
   orientation:
    a(b(x1)) = x1 + 12 >= x1 + 12 = b(a(x1))
    
    a(a(a(x1))) = x1 + 24 >= x1 + 24 = b(a(a(b(x1))))
    
    b(b(b(b(x1)))) = x1 + 16 >= x1 + 8 = a(x1)
   problem:
    a(b(x1)) -> b(a(x1))
    a(a(a(x1))) -> b(a(a(b(x1))))
   String Reversal Processor:
    b(a(x1)) -> a(b(x1))
    a(a(a(x1))) -> b(a(a(b(x1))))
    KBO Processor:
     weight function:
      w0 = 1
      w(a) = 1
      w(b) = 0
     precedence:
      b > a
     problem:
      
     Qed