YES

Problem:
 h(f(x),y) -> f(g(x,y))
 g(x,y) -> h(x,y)

Proof:
 DP Processor:
  DPs:
   h#(f(x),y) -> g#(x,y)
   g#(x,y) -> h#(x,y)
  TRS:
   h(f(x),y) -> f(g(x,y))
   g(x,y) -> h(x,y)
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [g#](x0, x1) = [0 2]x0 + [2],
    
    [h#](x0, x1) = [0 2]x0 + [1],
    
                  [2 0]     [2]
    [g](x0, x1) = [0 1]x0 + [2],
    
                  [2 0]     [2]
    [h](x0, x1) = [0 1]x0 + [2],
    
              [2 0]     [2]
    [f](x0) = [0 1]x0 + [1]
   orientation:
    h#(f(x),y) = [0 2]x + [3] >= [0 2]x + [2] = g#(x,y)
    
    g#(x,y) = [0 2]x + [2] >= [0 2]x + [1] = h#(x,y)
    
                [4 0]    [6]    [4 0]    [6]            
    h(f(x),y) = [0 1]x + [3] >= [0 1]x + [3] = f(g(x,y))
    
             [2 0]    [2]    [2 0]    [2]         
    g(x,y) = [0 1]x + [2] >= [0 1]x + [2] = h(x,y)
   problem:
    DPs:
     
    TRS:
     h(f(x),y) -> f(g(x,y))
     g(x,y) -> h(x,y)
   Qed