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

Proof:
 String Reversal Processor:
  a(a(x1)) -> b(b(b(x1)))
  a(x1) -> d(c(d(x1)))
  b(b(x1)) -> c(c(c(x1)))
  c(c(x1)) -> d(d(d(x1)))
  d(d(c(x1))) -> a(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c](x0) = x0 + 6,
    
    [d](x0) = x0 + 4,
    
    [b](x0) = x0 + 9,
    
    [a](x0) = x0 + 14
   orientation:
    a(a(x1)) = x1 + 28 >= x1 + 27 = b(b(b(x1)))
    
    a(x1) = x1 + 14 >= x1 + 14 = d(c(d(x1)))
    
    b(b(x1)) = x1 + 18 >= x1 + 18 = c(c(c(x1)))
    
    c(c(x1)) = x1 + 12 >= x1 + 12 = d(d(d(x1)))
    
    d(d(c(x1))) = x1 + 14 >= x1 + 14 = a(x1)
   problem:
    a(x1) -> d(c(d(x1)))
    b(b(x1)) -> c(c(c(x1)))
    c(c(x1)) -> d(d(d(x1)))
    d(d(c(x1))) -> a(x1)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]  
     [c](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]  
     [d](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]     [1]
     [b](x0) = [0 1 0]x0 + [0]
               [0 1 1]     [0],
     
               [1 0 0]  
     [a](x0) = [0 0 0]x0
               [0 0 0]  
    orientation:
             [1 0 0]      [1 0 0]                
     a(x1) = [0 0 0]x1 >= [0 0 0]x1 = d(c(d(x1)))
             [0 0 0]      [0 0 0]                
     
                [1 0 0]     [2]    [1 0 0]                
     b(b(x1)) = [0 1 0]x1 + [0] >= [0 0 0]x1 = c(c(c(x1)))
                [0 2 1]     [0]    [0 0 0]                
     
                [1 0 0]      [1 0 0]                
     c(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = d(d(d(x1)))
                [0 0 0]      [0 0 0]                
     
                   [1 0 0]      [1 0 0]          
     d(d(c(x1))) = [0 0 0]x1 >= [0 0 0]x1 = a(x1)
                   [0 0 0]      [0 0 0]          
    problem:
     a(x1) -> d(c(d(x1)))
     c(c(x1)) -> d(d(d(x1)))
     d(d(c(x1))) -> a(x1)
    String Reversal Processor:
     a(x1) -> d(c(d(x1)))
     c(c(x1)) -> d(d(d(x1)))
     c(d(d(x1))) -> a(x1)
     Bounds Processor:
      bound: 1
      enrichment: match
      automaton:
       final states: {7,5,1}
       transitions:
        a0(2) -> 7*
        d1(10) -> 11*
        d1(8) -> 9*
        c1(9) -> 10*
        f40() -> 2*
        d0(2) -> 3*
        d0(4) -> 1*
        d0(6) -> 5*
        d0(3) -> 6*
        c0(3) -> 4*
        2 -> 8*
        7 -> 10,4
        11 -> 7*
      problem:
       
      Qed