/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(c(c(c(x1))))
 b(c(x1)) -> d(d(d(d(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)) -> c(c(c(b(x1))))
  c(b(x1)) -> d(d(d(d(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:
    [d](x0) = x0 + 2,
    
    [b](x0) = x0 + 5,
    
    [c](x0) = x0 + 3,
    
    [a](x0) = x0 + 7
   orientation:
    a(a(x1)) = x1 + 14 >= x1 + 14 = c(c(c(b(x1))))
    
    c(b(x1)) = x1 + 8 >= x1 + 8 = d(d(d(d(x1))))
    
    a(x1) = x1 + 7 >= x1 + 7 = d(c(d(x1)))
    
    b(b(x1)) = x1 + 10 >= x1 + 9 = c(c(c(x1)))
    
    c(c(x1)) = x1 + 6 >= x1 + 6 = d(d(d(x1)))
    
    d(d(c(x1))) = x1 + 7 >= x1 + 7 = a(x1)
   problem:
    a(a(x1)) -> c(c(c(b(x1))))
    c(b(x1)) -> d(d(d(d(x1))))
    a(x1) -> d(c(d(x1)))
    c(c(x1)) -> d(d(d(x1)))
    d(d(c(x1))) -> a(x1)
   String Reversal Processor:
    a(a(x1)) -> b(c(c(c(x1))))
    b(c(x1)) -> d(d(d(d(x1))))
    a(x1) -> d(c(d(x1)))
    c(c(x1)) -> d(d(d(x1)))
    c(d(d(x1))) -> a(x1)
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [d](x0) = x0 + 2,
      
      [b](x0) = x0,
      
      [c](x0) = x0 + 8,
      
      [a](x0) = x0 + 12
     orientation:
      a(a(x1)) = x1 + 24 >= x1 + 24 = b(c(c(c(x1))))
      
      b(c(x1)) = x1 + 8 >= x1 + 8 = d(d(d(d(x1))))
      
      a(x1) = x1 + 12 >= x1 + 12 = d(c(d(x1)))
      
      c(c(x1)) = x1 + 16 >= x1 + 6 = d(d(d(x1)))
      
      c(d(d(x1))) = x1 + 12 >= x1 + 12 = a(x1)
     problem:
      a(a(x1)) -> b(c(c(c(x1))))
      b(c(x1)) -> d(d(d(d(x1))))
      a(x1) -> d(c(d(x1)))
      c(d(d(x1))) -> a(x1)
     String Reversal Processor:
      a(a(x1)) -> c(c(c(b(x1))))
      c(b(x1)) -> d(d(d(d(x1))))
      a(x1) -> d(c(d(x1)))
      d(d(c(x1))) -> a(x1)
      Matrix Interpretation Processor: dim=3
       
       interpretation:
                  [1 0 1]  
        [d](x0) = [0 0 0]x0
                  [1 0 0]  ,
        
                  [3 0 1]     [3]
        [b](x0) = [2 0 2]x0 + [0]
                  [3 1 2]     [1],
        
                  [1 1 0]     [1]
        [c](x0) = [0 0 0]x0 + [0]
                  [0 0 1]     [0],
        
                  [2 0 1]     [2]
        [a](x0) = [0 0 0]x0 + [0]
                  [1 1 1]     [1]
       orientation:
                   [5 1 3]     [7]    [5 0 3]     [6]                 
        a(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(c(c(b(x1))))
                   [3 1 2]     [4]    [3 1 2]     [1]                 
        
                   [5 0 3]     [4]    [5 0 3]                   
        c(b(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = d(d(d(d(x1))))
                   [3 1 2]     [1]    [3 0 2]                   
        
                [2 0 1]     [2]    [2 0 1]     [1]              
        a(x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = d(c(d(x1)))
                [1 1 1]     [1]    [1 0 1]     [1]              
        
                      [2 2 1]     [2]    [2 0 1]     [2]        
        d(d(c(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(x1)
                      [1 1 1]     [1]    [1 1 1]     [1]        
       problem:
        d(d(c(x1))) -> a(x1)
       String Reversal Processor:
        c(d(d(x1))) -> a(x1)
        KBO Processor:
         weight function:
          w0 = 1
          w(d) = w(c) = 1
          w(a) = 0
         precedence:
          a > d ~ c
         problem:
          
         Qed