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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [b](x0) = [0 0 0]x0
             [0 0 0]  ,
   
             [1 0 1]     [0]
   [a](x0) = [0 0 1]x0 + [1]
             [0 1 0]     [0]
  orientation:
           [1 0 1]     [0]    [1 0 0]          
   a(x1) = [0 0 1]x1 + [1] >= [0 0 0]x1 = b(x1)
           [0 1 0]     [0]    [0 0 0]          
   
              [1 1 1]     [0]    [1 0 1]     [0]              
   a(a(x1)) = [0 1 0]x1 + [1] >= [0 0 0]x1 + [1] = a(b(a(x1)))
              [0 0 1]     [1]    [0 0 0]     [0]              
   
              [1 0 0]     [0]    [1 0 0]                
   a(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 = b(b(b(x1)))
              [0 0 0]     [0]    [0 0 0]                
   
                 [1 1 2]     [1]    [1 1 1]     [0]                    
   a(a(a(x1))) = [0 0 1]x1 + [2] >= [0 0 0]x1 + [1] = a(a(b(a(a(x1)))))
                 [0 1 0]     [1]    [0 0 0]     [1]                    
   
                 [1 0 0]     [0]    [1 0 0]     [0]                    
   a(a(b(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(b(b(a(b(x1)))))
                 [0 0 0]     [1]    [0 0 0]     [0]                    
   
                 [1 0 1]     [0]    [1 0 1]                      
   a(b(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = b(a(b(b(a(x1)))))
                 [0 0 0]     [0]    [0 0 0]                      
   
                 [1 0 0]     [0]    [1 0 0]                      
   a(b(b(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = b(b(b(b(b(x1)))))
                 [0 0 0]     [0]    [0 0 0]                      
   
              [1 0 1]      [1 0 0]                
   b(a(x1)) = [0 0 0]x1 >= [0 0 0]x1 = b(b(b(x1)))
              [0 0 0]      [0 0 0]                
   
                 [1 0 1]     [0]    [1 0 0]     [0]                    
   a(b(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(b(b(a(b(x1)))))
                 [0 0 0]     [0]    [0 0 0]     [0]                    
   
                 [1 1 1]      [1 0 1]                      
   b(a(a(x1))) = [0 0 0]x1 >= [0 0 0]x1 = b(a(b(b(a(x1)))))
                 [0 0 0]      [0 0 0]                      
   
                 [1 0 1]      [1 0 0]                      
   b(b(a(x1))) = [0 0 0]x1 >= [0 0 0]x1 = b(b(b(b(b(x1)))))
                 [0 0 0]      [0 0 0]                      
  problem:
   a(x1) -> b(x1)
   a(a(x1)) -> a(b(a(x1)))
   a(b(x1)) -> b(b(b(x1)))
   a(a(b(x1))) -> a(b(b(a(b(x1)))))
   a(b(a(x1))) -> b(a(b(b(a(x1)))))
   a(b(b(x1))) -> b(b(b(b(b(x1)))))
   b(a(x1)) -> b(b(b(x1)))
   a(b(a(x1))) -> a(b(b(a(b(x1)))))
   b(a(a(x1))) -> b(a(b(b(a(x1)))))
   b(b(a(x1))) -> b(b(b(b(b(x1)))))
  String Reversal Processor:
   a(x1) -> b(x1)
   a(a(x1)) -> a(b(a(x1)))
   b(a(x1)) -> b(b(b(x1)))
   b(a(a(x1))) -> b(a(b(b(a(x1)))))
   a(b(a(x1))) -> a(b(b(a(b(x1)))))
   b(b(a(x1))) -> b(b(b(b(b(x1)))))
   a(b(x1)) -> b(b(b(x1)))
   a(b(a(x1))) -> b(a(b(b(a(x1)))))
   a(a(b(x1))) -> a(b(b(a(b(x1)))))
   a(b(b(x1))) -> b(b(b(b(b(x1)))))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [b](x0) = x0,
     
     [a](x0) = x0 + 4
    orientation:
     a(x1) = x1 + 4 >= x1 = b(x1)
     
     a(a(x1)) = x1 + 8 >= x1 + 8 = a(b(a(x1)))
     
     b(a(x1)) = x1 + 4 >= x1 = b(b(b(x1)))
     
     b(a(a(x1))) = x1 + 8 >= x1 + 8 = b(a(b(b(a(x1)))))
     
     a(b(a(x1))) = x1 + 8 >= x1 + 8 = a(b(b(a(b(x1)))))
     
     b(b(a(x1))) = x1 + 4 >= x1 = b(b(b(b(b(x1)))))
     
     a(b(x1)) = x1 + 4 >= x1 = b(b(b(x1)))
     
     a(b(a(x1))) = x1 + 8 >= x1 + 8 = b(a(b(b(a(x1)))))
     
     a(a(b(x1))) = x1 + 8 >= x1 + 8 = a(b(b(a(b(x1)))))
     
     a(b(b(x1))) = x1 + 4 >= x1 = b(b(b(b(b(x1)))))
    problem:
     a(a(x1)) -> a(b(a(x1)))
     b(a(a(x1))) -> b(a(b(b(a(x1)))))
     a(b(a(x1))) -> a(b(b(a(b(x1)))))
     a(b(a(x1))) -> b(a(b(b(a(x1)))))
     a(a(b(x1))) -> a(b(b(a(b(x1)))))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {8,5,1}
      transitions:
       b1(19) -> 20*
       b1(27) -> 28*
       b1(43) -> 44*
       b1(37) -> 38*
       b1(21) -> 22*
       b1(55) -> 56*
       b1(45) -> 46*
       b1(29) -> 30*
       b1(26) -> 27*
       b1(22) -> 23*
       f20() -> 2*
       a0(6) -> 7*
       a0(2) -> 3*
       a0(4) -> 1*
       a0(9) -> 10*
       a0(12) -> 8*
       b0(3) -> 4*
       b0(11) -> 12*
       b0(4) -> 6*
       b0(2) -> 9*
       b0(10) -> 11*
       b0(7) -> 5*
       a1(20) -> 21*
       a1(25) -> 26*
       a1(57) -> 58*
       a1(23) -> 24*
       a1(51) -> 52*
       a1(28) -> 29*
       56 -> 20*
       24 -> 21,10,52,26,45,1
       46 -> 27*
       30 -> 21,10,52,26,1
       4 -> 51,43
       2 -> 25,19
       52 -> 26*
       6 -> 57,55
       58 -> 26*
       38 -> 20*
       44 -> 20*
       1 -> 3*
       5 -> 20,9,4
       23 -> 37*
     problem:
      
     Qed