YES

Problem:
 c(c(c(a(x,y)))) -> b(c(c(c(c(y)))),x)
 c(c(b(c(y),0()))) -> a(0(),c(c(a(y,0()))))
 c(c(a(a(y,0()),x))) -> c(y)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = 2x0,
   
   [0] = 1,
   
   [a](x0, x1) = x0 + 2x1,
   
   [b](x0, x1) = x0 + 6x1
  orientation:
   c(c(c(a(x,y)))) = 8x + 16y >= 6x + 16y = b(c(c(c(c(y)))),x)
   
   c(c(b(c(y),0()))) = 8y + 24 >= 8y + 17 = a(0(),c(c(a(y,0()))))
   
   c(c(a(a(y,0()),x))) = 8x + 4y + 8 >= 2y = c(y)
  problem:
   c(c(c(a(x,y)))) -> b(c(c(c(c(y)))),x)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c](x0) = x0,
    
    [a](x0, x1) = x0 + x1 + 3,
    
    [b](x0, x1) = x0 + x1
   orientation:
    c(c(c(a(x,y)))) = x + y + 3 >= x + y = b(c(c(c(c(y)))),x)
   problem:
    
   Qed