YES

Problem:
 f(j(x,y),y) -> g(f(x,k(y)))
 f(x,h1(y,z)) -> h2(0(),x,h1(y,z))
 g(h2(x,y,h1(z,u))) -> h2(s(x),y,h1(z,u))
 h2(x,j(y,h1(z,u)),h1(z,u)) -> h2(s(x),y,h1(s(z),u))
 i(f(x,h(y))) -> y
 i(h2(s(x),y,h1(x,z))) -> z
 k(h(x)) -> h1(0(),x)
 k(h1(x,y)) -> h1(s(x),y)

Proof:
 DP Processor:
  DPs:
   f#(j(x,y),y) -> k#(y)
   f#(j(x,y),y) -> f#(x,k(y))
   f#(j(x,y),y) -> g#(f(x,k(y)))
   f#(x,h1(y,z)) -> h2#(0(),x,h1(y,z))
   g#(h2(x,y,h1(z,u))) -> h2#(s(x),y,h1(z,u))
   h2#(x,j(y,h1(z,u)),h1(z,u)) -> h2#(s(x),y,h1(s(z),u))
  TRS:
   f(j(x,y),y) -> g(f(x,k(y)))
   f(x,h1(y,z)) -> h2(0(),x,h1(y,z))
   g(h2(x,y,h1(z,u))) -> h2(s(x),y,h1(z,u))
   h2(x,j(y,h1(z,u)),h1(z,u)) -> h2(s(x),y,h1(s(z),u))
   i(f(x,h(y))) -> y
   i(h2(s(x),y,h1(x,z))) -> z
   k(h(x)) -> h1(0(),x)
   k(h1(x,y)) -> h1(s(x),y)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [h2#](x0, x1, x2) = x1 + x2,
    
    [g#](x0) = x0 + 2,
    
    [k#](x0) = 1x0,
    
    [f#](x0, x1) = 4x0 + 4x1 + 0,
    
    [i](x0) = x0 + 8,
    
    [h](x0) = x0 + 0,
    
    [s](x0) = x0 + 6,
    
    [h2](x0, x1, x2) = 4x1 + 1x2,
    
    [0] = 6,
    
    [h1](x0, x1) = x1,
    
    [g](x0) = 1x0 + 0,
    
    [k](x0) = 1x0 + 0,
    
    [f](x0, x1) = 5x0 + 4x1 + 0,
    
    [j](x0, x1) = 2x0 + 2x1 + 1
   orientation:
    f#(j(x,y),y) = 6x + 6y + 5 >= 1y = k#(y)
    
    f#(j(x,y),y) = 6x + 6y + 5 >= 4x + 5y + 4 = f#(x,k(y))
    
    f#(j(x,y),y) = 6x + 6y + 5 >= 5x + 5y + 4 = g#(f(x,k(y)))
    
    f#(x,h1(y,z)) = 4x + 4z + 0 >= x + z = h2#(0(),x,h1(y,z))
    
    g#(h2(x,y,h1(z,u))) = 1u + 4y + 2 >= u + y = h2#(s(x),y,h1(z,u))
    
    h2#(x,j(y,h1(z,u)),h1(z,u)) = 2u + 2y + 1 >= u + y = h2#(s(x),y,h1(s(z),u))
    
    f(j(x,y),y) = 7x + 7y + 6 >= 6x + 6y + 5 = g(f(x,k(y)))
    
    f(x,h1(y,z)) = 5x + 4z + 0 >= 4x + 1z = h2(0(),x,h1(y,z))
    
    g(h2(x,y,h1(z,u))) = 2u + 5y + 0 >= 1u + 4y = h2(s(x),y,h1(z,u))
    
    h2(x,j(y,h1(z,u)),h1(z,u)) = 6u + 6y + 5 >= 1u + 4y = h2(s(x),y,h1(s(z),u))
    
    i(f(x,h(y))) = 5x + 4y + 8 >= y = y
    
    i(h2(s(x),y,h1(x,z))) = 4y + 1z + 8 >= z = z
    
    k(h(x)) = 1x + 1 >= x = h1(0(),x)
    
    k(h1(x,y)) = 1y + 0 >= y = h1(s(x),y)
   problem:
    DPs:
     
    TRS:
     f(j(x,y),y) -> g(f(x,k(y)))
     f(x,h1(y,z)) -> h2(0(),x,h1(y,z))
     g(h2(x,y,h1(z,u))) -> h2(s(x),y,h1(z,u))
     h2(x,j(y,h1(z,u)),h1(z,u)) -> h2(s(x),y,h1(s(z),u))
     i(f(x,h(y))) -> y
     i(h2(s(x),y,h1(x,z))) -> z
     k(h(x)) -> h1(0(),x)
     k(h1(x,y)) -> h1(s(x),y)
   Qed