YES

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

Proof:
 DP Processor:
  DPs:
   b#(x,y) -> a#(0(),x)
   b#(x,y) -> a#(c(y),a(0(),x))
   a#(y,c(b(a(0(),x),0()))) -> b#(0(),y)
   a#(y,c(b(a(0(),x),0()))) -> a#(c(b(0(),y)),x)
   a#(y,c(b(a(0(),x),0()))) -> b#(a(c(b(0(),y)),x),0())
  TRS:
   b(x,y) -> c(a(c(y),a(0(),x)))
   a(y,x) -> y
   a(y,c(b(a(0(),x),0()))) -> b(a(c(b(0(),y)),x),0())
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [a#](x0, x1) = [1 0]x1,
    
    [b#](x0, x1) = [1 0]x0 + [1],
    
                  [1 0]     [0 0]  
    [a](x0, x1) = [0 2]x0 + [1 0]x1,
    
          [0]
    [0] = [1],
    
              [0 2]  
    [c](x0) = [0 0]x0,
    
                  [0 0]     [0]
    [b](x0, x1) = [0 1]x0 + [1]
   orientation:
    b#(x,y) = [1 0]x + [1] >= [1 0]x = a#(0(),x)
    
    b#(x,y) = [1 0]x + [1] >= [0] = a#(c(y),a(0(),x))
    
    a#(y,c(b(a(0(),x),0()))) = [2 0]x + [6] >= [1] = b#(0(),y)
    
    a#(y,c(b(a(0(),x),0()))) = [2 0]x + [6] >= [1 0]x = a#(c(b(0(),y)),x)
    
    a#(y,c(b(a(0(),x),0()))) = [2 0]x + [6] >= [5] = b#(a(c(b(0(),y)),x),0())
    
             [0 0]    [0]    [0]                      
    b(x,y) = [0 1]x + [1] >= [0] = c(a(c(y),a(0(),x)))
    
             [0 0]    [1 0]          
    a(y,x) = [1 0]x + [0 2]y >= y = y
    
                              [0 0]    [1 0]    [0]    [0 0]    [0]                          
    a(y,c(b(a(0(),x),0()))) = [2 0]x + [0 2]y + [6] >= [1 0]x + [1] = b(a(c(b(0(),y)),x),0())
   problem:
    DPs:
     
    TRS:
     b(x,y) -> c(a(c(y),a(0(),x)))
     a(y,x) -> y
     a(y,c(b(a(0(),x),0()))) -> b(a(c(b(0(),y)),x),0())
   Qed