YES

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

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