YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [c](x0) = [0 1 0]x0
             [0 0 0]  ,
   
             [1 0 0]  
   [b](x0) = [0 0 1]x0
             [0 0 1]  ,
   
             [1 0 0]  
   [d](x0) = [0 0 0]x0
             [0 0 1]  ,
   
             [1 1 0]     [0]
   [a](x0) = [0 0 0]x0 + [0]
             [0 0 1]     [1]
  orientation:
                             [1 1 1]     [1]    [1 0 1]     [0]                          
   a(a(b(d(b(d(a(x1))))))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(a(c(a(a(b(d(x1)))))))
                             [0 0 1]     [3]    [0 0 0]     [2]                          
   
                 [1 1 0]     [0]    [1 1 0]                   
   a(a(c(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 = c(c(a(a(x1))))
                 [0 0 0]     [2]    [0 0 0]                   
   
                 [1 0 0]      [1 0 0]                      
   c(c(c(x1))) = [0 1 0]x1 >= [0 0 0]x1 = b(d(c(b(d(x1)))))
                 [0 0 0]      [0 0 0]                      
  problem:
   a(a(c(x1))) -> c(c(a(a(x1))))
   c(c(c(x1))) -> b(d(c(b(d(x1)))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c](x0) = x0 + 1,
    
    [b](x0) = x0,
    
    [d](x0) = x0 + 1,
    
    [a](x0) = 4x0
   orientation:
    a(a(c(x1))) = 16x1 + 16 >= 16x1 + 2 = c(c(a(a(x1))))
    
    c(c(c(x1))) = x1 + 3 >= x1 + 3 = b(d(c(b(d(x1)))))
   problem:
    c(c(c(x1))) -> b(d(c(b(d(x1)))))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 0]  
     [c](x0) = [2 0]x0,
     
               [0  0 ]  
     [b](x0) = [-& -&]x0,
     
               [0  -&]  
     [d](x0) = [0  -&]x0
    orientation:
                   [2 2]      [0  -&]                      
     c(c(c(x1))) = [4 2]x1 >= [-& -&]x1 = b(d(c(b(d(x1)))))
    problem:
     
    Qed