YES

Problem:
 f(x,y) -> g1(x,x,y)
 f(x,y) -> g1(y,x,x)
 f(x,y) -> g2(x,y,y)
 f(x,y) -> g2(y,y,x)
 g1(x,x,y) -> h(x,y)
 g1(y,x,x) -> h(x,y)
 g2(x,y,y) -> h(x,y)
 g2(y,y,x) -> h(x,y)
 h(x,x) -> x

Proof:
 Polynomial Interpretation Processor:
  dimension: 1
  interpretation:
   [h](x0, x1) = x0 + x1 + 2,
   
   [g2](x0, x1, x2) = 3x0 + 2x1 + 3x2 + 4,
   
   [g1](x0, x1, x2) = x0 + 2x1 + x2 + 4,
   
   [f](x0, x1) = 3x0 + 5x1 + 5
  orientation:
   f(x,y) = 3x + 5y + 5 >= 3x + y + 4 = g1(x,x,y)
   
   f(x,y) = 3x + 5y + 5 >= 3x + y + 4 = g1(y,x,x)
   
   f(x,y) = 3x + 5y + 5 >= 3x + 5y + 4 = g2(x,y,y)
   
   f(x,y) = 3x + 5y + 5 >= 3x + 5y + 4 = g2(y,y,x)
   
   g1(x,x,y) = 3x + y + 4 >= x + y + 2 = h(x,y)
   
   g1(y,x,x) = 3x + y + 4 >= x + y + 2 = h(x,y)
   
   g2(x,y,y) = 3x + 5y + 4 >= x + y + 2 = h(x,y)
   
   g2(y,y,x) = 3x + 5y + 4 >= x + y + 2 = h(x,y)
   
   h(x,x) = 2x + 2 >= x = x
  problem:
   
  Qed