/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

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

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