/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

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

Proof:
 String Reversal Processor:
  a(c(a(x1))) -> c(a(c(x1)))
  b(a(a(x1))) -> b(d(a(x1)))
  b(a(x1)) -> a(a(b(x1)))
  d(d(x1)) -> b(d(a(x1)))
  b(b(x1)) -> c(b(x1))
  c(d(a(x1))) -> a(c(x1))
  c(b(x1)) -> a(a(a(x1)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
              [1 2]     [0]
    [d](x0) = [0 0]x0 + [3],
    
              [1 0]     [0]
    [b](x0) = [0 0]x0 + [3],
    
              [1 0]  
    [c](x0) = [0 0]x0,
    
              [1 0]  
    [a](x0) = [0 0]x0
   orientation:
                  [1 0]      [1 0]                
    a(c(a(x1))) = [0 0]x1 >= [0 0]x1 = c(a(c(x1)))
    
                  [1 0]     [0]    [1 0]     [0]              
    b(a(a(x1))) = [0 0]x1 + [3] >= [0 0]x1 + [3] = b(d(a(x1)))
    
               [1 0]     [0]    [1 0]                
    b(a(x1)) = [0 0]x1 + [3] >= [0 0]x1 = a(a(b(x1)))
    
               [1 2]     [6]    [1 0]     [0]              
    d(d(x1)) = [0 0]x1 + [3] >= [0 0]x1 + [3] = b(d(a(x1)))
    
               [1 0]     [0]    [1 0]             
    b(b(x1)) = [0 0]x1 + [3] >= [0 0]x1 = c(b(x1))
    
                  [1 0]      [1 0]             
    c(d(a(x1))) = [0 0]x1 >= [0 0]x1 = a(c(x1))
    
               [1 0]      [1 0]                
    c(b(x1)) = [0 0]x1 >= [0 0]x1 = a(a(a(x1)))
   problem:
    a(c(a(x1))) -> c(a(c(x1)))
    b(a(a(x1))) -> b(d(a(x1)))
    b(a(x1)) -> a(a(b(x1)))
    b(b(x1)) -> c(b(x1))
    c(d(a(x1))) -> a(c(x1))
    c(b(x1)) -> a(a(a(x1)))
   String Reversal Processor:
    a(c(a(x1))) -> c(a(c(x1)))
    a(a(b(x1))) -> a(d(b(x1)))
    a(b(x1)) -> b(a(a(x1)))
    b(b(x1)) -> b(c(x1))
    a(d(c(x1))) -> c(a(x1))
    b(c(x1)) -> a(a(a(x1)))
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [d](x0) = x0,
      
      [b](x0) = x0 + 1,
      
      [c](x0) = x0,
      
      [a](x0) = x0
     orientation:
      a(c(a(x1))) = x1 >= x1 = c(a(c(x1)))
      
      a(a(b(x1))) = x1 + 1 >= x1 + 1 = a(d(b(x1)))
      
      a(b(x1)) = x1 + 1 >= x1 + 1 = b(a(a(x1)))
      
      b(b(x1)) = x1 + 2 >= x1 + 1 = b(c(x1))
      
      a(d(c(x1))) = x1 >= x1 = c(a(x1))
      
      b(c(x1)) = x1 + 1 >= x1 = a(a(a(x1)))
     problem:
      a(c(a(x1))) -> c(a(c(x1)))
      a(a(b(x1))) -> a(d(b(x1)))
      a(b(x1)) -> b(a(a(x1)))
      a(d(c(x1))) -> c(a(x1))
     String Reversal Processor:
      a(c(a(x1))) -> c(a(c(x1)))
      b(a(a(x1))) -> b(d(a(x1)))
      b(a(x1)) -> a(a(b(x1)))
      c(d(a(x1))) -> a(c(x1))
      Bounds Processor:
       bound: 1
       enrichment: match
       automaton:
        final states: {4,8,5,1}
        transitions:
         c1(15) -> 16*
         c1(17) -> 18*
         c1(21) -> 22*
         a1(16) -> 17*
         f40() -> 2*
         c0(2) -> 3*
         c0(4) -> 1*
         a0(10) -> 8*
         a0(2) -> 6*
         a0(9) -> 10*
         a0(3) -> 4*
         b0(7) -> 5*
         b0(2) -> 9*
         d0(6) -> 7*
         1 -> 6,4
         3 -> 15*
         4 -> 3*
         5 -> 9*
         8 -> 9*
         16 -> 21*
         18 -> 17,4,3
         22 -> 16*
       problem:
        
       Qed