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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]     [0]
   [f](x0) = [0 0 0]x0 + [0]
             [0 0 0]     [1],
   
             [1 0 0]     [0]
   [c](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 0]     [0]
   [d](x0) = [0 0 0]x0 + [1]
             [0 1 0]     [0],
   
             [1 0 1]     [0]
   [g](x0) = [0 0 1]x0 + [0]
             [0 0 0]     [1],
   
             [1 0 0]     [0]
   [a](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 0]     [0]
   [b](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0]
  orientation:
              [1 0 0]     [0]    [1 0 0]     [0]           
   a(a(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(c(x1))
              [0 0 0]     [0]    [0 0 0]     [0]           
   
              [1 0 0]     [0]    [1 0 0]     [0]           
   b(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = c(d(x1))
              [0 0 0]     [0]    [0 0 0]     [0]           
   
           [1 0 0]     [0]    [1 0 0]     [0]        
   b(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(x1)
           [0 0 0]     [0]    [0 0 0]     [0]        
   
              [1 0 0]     [0]    [1 0 0]     [0]           
   c(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(f(x1))
              [0 0 0]     [0]    [0 0 0]     [0]           
   
              [1 0 0]     [0]    [1 0 0]     [0]              
   d(d(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = f(f(f(x1)))
              [0 0 0]     [1]    [0 0 0]     [1]              
   
           [1 0 0]     [0]    [1 0 0]     [0]        
   d(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(x1)
           [0 1 0]     [0]    [0 0 0]     [0]        
   
              [1 0 0]     [0]    [1 0 0]     [0]           
   f(f(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = g(a(x1))
              [0 0 0]     [1]    [0 0 0]     [1]           
   
              [1 0 1]     [1]    [1 0 0]     [0]        
   g(g(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(x1)
              [0 0 0]     [1]    [0 0 0]     [0]        
  problem:
   a(a(x1)) -> b(c(x1))
   b(b(x1)) -> c(d(x1))
   b(x1) -> a(x1)
   c(c(x1)) -> d(f(x1))
   d(d(x1)) -> f(f(f(x1)))
   d(x1) -> b(x1)
   f(f(x1)) -> g(a(x1))
  String Reversal Processor:
   a(a(x1)) -> c(b(x1))
   b(b(x1)) -> d(c(x1))
   b(x1) -> a(x1)
   c(c(x1)) -> f(d(x1))
   d(d(x1)) -> f(f(f(x1)))
   d(x1) -> b(x1)
   f(f(x1)) -> a(g(x1))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [f](x0) = x0 + 6,
     
     [c](x0) = x0 + 9,
     
     [d](x0) = x0 + 12,
     
     [g](x0) = x0,
     
     [a](x0) = x0 + 12,
     
     [b](x0) = x0 + 12
    orientation:
     a(a(x1)) = x1 + 24 >= x1 + 21 = c(b(x1))
     
     b(b(x1)) = x1 + 24 >= x1 + 21 = d(c(x1))
     
     b(x1) = x1 + 12 >= x1 + 12 = a(x1)
     
     c(c(x1)) = x1 + 18 >= x1 + 18 = f(d(x1))
     
     d(d(x1)) = x1 + 24 >= x1 + 18 = f(f(f(x1)))
     
     d(x1) = x1 + 12 >= x1 + 12 = b(x1)
     
     f(f(x1)) = x1 + 12 >= x1 + 12 = a(g(x1))
    problem:
     b(x1) -> a(x1)
     c(c(x1)) -> f(d(x1))
     d(x1) -> b(x1)
     f(f(x1)) -> a(g(x1))
    KBO Processor:
     weight function:
      w0 = 1
      w(g) = w(f) = w(d) = w(b) = w(c) = w(a) = 1
     precedence:
      c > f > d > b > a > g
     problem:
      
     Qed