YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [c](x0) = 4x0,
   
   [b](x0) = 3x0,
   
   [g](x0) = 2x0,
   
   [d](x0) = x0,
   
   [a](x0) = 2x0
  orientation:
   a(x1) = 2x1 >= 2x1 = g(d(x1))
   
   b(b(b(x1))) = 9x1 >= 8x1 = c(d(c(x1)))
   
   b(b(x1)) = 6x1 >= 6x1 = a(g(g(x1)))
   
   c(d(x1)) = 4x1 >= 4x1 = g(g(x1))
   
   g(g(g(x1))) = 6x1 >= 6x1 = b(b(x1))
  problem:
   a(x1) -> g(d(x1))
   b(b(x1)) -> a(g(g(x1)))
   c(d(x1)) -> g(g(x1))
   g(g(g(x1))) -> b(b(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [3 1]  
    [c](x0) = [3 2]x0,
    
              [1 2]  
    [b](x0) = [0 0]x0,
    
              [0 1]  
    [g](x0) = [0 1]x0,
    
              [0  0 ]  
    [d](x0) = [-& -&]x0,
    
              [1 1]  
    [a](x0) = [0 0]x0
   orientation:
            [1 1]      [0 0]             
    a(x1) = [0 0]x1 >= [0 0]x1 = g(d(x1))
    
               [2 3]      [2 3]                
    b(b(x1)) = [1 2]x1 >= [1 2]x1 = a(g(g(x1)))
    
               [3 3]      [1 2]             
    c(d(x1)) = [3 3]x1 >= [1 2]x1 = g(g(x1))
    
                  [2 3]      [2 3]             
    g(g(g(x1))) = [2 3]x1 >= [1 2]x1 = b(b(x1))
   problem:
    a(x1) -> g(d(x1))
    b(b(x1)) -> a(g(g(x1)))
    g(g(g(x1))) -> b(b(x1))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 1]  
     [b](x0) = [0 1]x0,
     
               [0 0]  
     [g](x0) = [0 1]x0,
     
               [0  -&]  
     [d](x0) = [-& -&]x0,
     
               [0  -&]  
     [a](x0) = [0  -&]x0
    orientation:
             [0  -&]      [0  -&]             
     a(x1) = [0  -&]x1 >= [0  -&]x1 = g(d(x1))
     
                [1 2]      [0 1]                
     b(b(x1)) = [1 2]x1 >= [0 1]x1 = a(g(g(x1)))
     
                   [1 2]      [1 2]             
     g(g(g(x1))) = [2 3]x1 >= [1 2]x1 = b(b(x1))
    problem:
     a(x1) -> g(d(x1))
     g(g(g(x1))) -> b(b(x1))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
                [0 0]  
      [b](x0) = [0 0]x0,
      
                [0 0]  
      [g](x0) = [0 1]x0,
      
                [1 0]  
      [d](x0) = [0 1]x0,
      
                [1 2]  
      [a](x0) = [2 3]x0
     orientation:
              [1 2]      [1 1]             
      a(x1) = [2 3]x1 >= [1 2]x1 = g(d(x1))
      
                    [1 2]      [0 0]             
      g(g(g(x1))) = [2 3]x1 >= [0 0]x1 = b(b(x1))
     problem:
      a(x1) -> g(d(x1))
     Arctic Interpretation Processor:
      dimension: 3
      interpretation:
                 [0  0  1 ]  
       [g](x0) = [-& -& 0 ]x0
                 [0  2  1 ]  ,
       
                 [0  2  0 ]  
       [d](x0) = [0  0  -&]x0
                 [0  1  0 ]  ,
       
                 [2 3 2]  
       [a](x0) = [2 2 2]x0
                 [3 3 2]  
      orientation:
               [2 3 2]      [1 2 1]             
       a(x1) = [2 2 2]x1 >= [0 1 0]x1 = g(d(x1))
               [3 3 2]      [2 2 1]             
      problem:
       
      Qed