YES

Problem:
 f(c(a(),z,x)) -> b(a(),z)
 b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y)))
 b(y,z) -> z

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [b](x0, x1) = 4x0 + 2x1 + 4,
   
   [f](x0) = x0,
   
   [c](x0, x1, x2) = x0 + 2x1 + x2 + 4,
   
   [a] = 0
  orientation:
   f(c(a(),z,x)) = x + 2z + 4 >= 2z + 4 = b(a(),z)
   
   b(x,b(z,y)) = 4x + 4y + 8z + 12 >= 2x + 2y + 8z + 12 = f(b(f(f(z)),c(x,z,y)))
   
   b(y,z) = 4y + 2z + 4 >= z = z
  problem:
   f(c(a(),z,x)) -> b(a(),z)
   b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
                  [2 0]     [2 2]     [0]
    [b](x0, x1) = [1 2]x0 + [0 0]x1 + [2],
    
              [1 0]     [0]
    [f](x0) = [0 0]x0 + [2],
    
                      [1 0]     [2 2]     [1 2]     [1]
    [c](x0, x1, x2) = [0 0]x0 + [0 0]x1 + [1 0]x2 + [1],
    
          [0]
    [a] = [0]
   orientation:
                    [1 2]    [2 2]    [1]    [2 2]    [0]           
    f(c(a(),z,x)) = [0 0]x + [0 0]z + [2] >= [0 0]z + [2] = b(a(),z)
    
                  [2 0]    [4 4]    [6 4]    [4]    [2 0]    [4 4]    [6 4]    [4]                         
    b(x,b(z,y)) = [1 2]x + [0 0]y + [0 0]z + [2] >= [0 0]x + [0 0]y + [0 0]z + [2] = f(b(f(f(z)),c(x,z,y)))
   problem:
    b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y)))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [b](x0, x1) = 2x0 + 2x1 + 5,
     
     [f](x0) = x0 + 1,
     
     [c](x0, x1, x2) = x0 + x1 + 2x2
    orientation:
     b(x,b(z,y)) = 2x + 4y + 4z + 15 >= 2x + 4y + 4z + 10 = f(b(f(f(z)),c(x,z,y)))
    problem:
     
    Qed