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)
  Usable Rule Processor:
   DPs:
    h#(f(x),y) -> g#(x,y)
    g#(x,y) -> h#(x,y)
   TRS:
    
   EDG Processor:
    DPs:
     h#(f(x),y) -> g#(x,y)
     g#(x,y) -> h#(x,y)
    TRS:
     
    graph:
     g#(x,y) -> h#(x,y) -> h#(f(x),y) -> g#(x,y)
     h#(f(x),y) -> g#(x,y) -> g#(x,y) -> h#(x,y)
    Restore Modifier:
     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)
     SCC Processor:
      #sccs: 1
      #rules: 2
      #arcs: 2/4
      DPs:
       g#(x,y) -> h#(x,y)
       h#(f(x),y) -> g#(x,y)
      TRS:
       h(f(x),y) -> f(g(x,y))
       g(x,y) -> h(x,y)
      Matrix Interpretation Processor:
       dimension: 1
       interpretation:
        [g#](x0, x1) = x0 + 1,
        
        [h#](x0, x1) = x0,
        
        [g](x0, x1) = x0,
        
        [h](x0, x1) = x0,
        
        [f](x0) = x0 + 1
       orientation:
        g#(x,y) = x + 1 >= x = h#(x,y)
        
        h#(f(x),y) = x + 1 >= x + 1 = g#(x,y)
        
        h(f(x),y) = x + 1 >= x + 1 = f(g(x,y))
        
        g(x,y) = x >= x = h(x,y)
       problem:
        DPs:
         h#(f(x),y) -> g#(x,y)
        TRS:
         h(f(x),y) -> f(g(x,y))
         g(x,y) -> h(x,y)
       Matrix Interpretation Processor:
        dimension: 1
        interpretation:
         [g#](x0, x1) = 0,
         
         [h#](x0, x1) = x0 + 1,
         
         [g](x0, x1) = x0,
         
         [h](x0, x1) = x0,
         
         [f](x0) = 1
        orientation:
         h#(f(x),y) = 2 >= 0 = g#(x,y)
         
         h(f(x),y) = 1 >= 1 = f(g(x,y))
         
         g(x,y) = x >= x = h(x,y)
        problem:
         DPs:
          
         TRS:
          h(f(x),y) -> f(g(x,y))
          g(x,y) -> h(x,y)
        Qed