YES

We show the termination of the TRS R:

  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)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(j(x,y),y) -> g#(f(x,k(y)))
p2: f#(j(x,y),y) -> f#(x,k(y))
p3: f#(j(x,y),y) -> k#(y)
p4: f#(x,h1(y,z)) -> h2#(|0|(),x,h1(y,z))
p5: g#(h2(x,y,h1(z,u))) -> h2#(s(x),y,h1(z,u))
p6: h2#(x,j(y,h1(z,u)),h1(z,u)) -> h2#(s(x),y,h1(s(z),u))

and R consists of:

r1: f(j(x,y),y) -> g(f(x,k(y)))
r2: f(x,h1(y,z)) -> h2(|0|(),x,h1(y,z))
r3: g(h2(x,y,h1(z,u))) -> h2(s(x),y,h1(z,u))
r4: h2(x,j(y,h1(z,u)),h1(z,u)) -> h2(s(x),y,h1(s(z),u))
r5: i(f(x,h(y))) -> y
r6: i(h2(s(x),y,h1(x,z))) -> z
r7: k(h(x)) -> h1(|0|(),x)
r8: k(h1(x,y)) -> h1(s(x),y)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p6}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(j(x,y),y) -> f#(x,k(y))

and R consists of:

r1: f(j(x,y),y) -> g(f(x,k(y)))
r2: f(x,h1(y,z)) -> h2(|0|(),x,h1(y,z))
r3: g(h2(x,y,h1(z,u))) -> h2(s(x),y,h1(z,u))
r4: h2(x,j(y,h1(z,u)),h1(z,u)) -> h2(s(x),y,h1(s(z),u))
r5: i(f(x,h(y))) -> y
r6: i(h2(s(x),y,h1(x,z))) -> z
r7: k(h(x)) -> h1(|0|(),x)
r8: k(h1(x,y)) -> h1(s(x),y)

The set of usable rules consists of

  r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2
        j_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (3,1)
        k_A(x1) = ((1,0),(1,1)) x1 + (2,1)
        h_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        h1_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        |0|_A() = (1,1)
        s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = x1 + x2
        j_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (1,1)
        k_A(x1) = ((0,1),(1,1)) x1 + (1,1)
        h_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        h1_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,1),(0,1)) x2 + (3,3)
        |0|_A() = (1,1)
        s_A(x1) = ((1,1),(0,0)) x1 + (5,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: h2#(x,j(y,h1(z,u)),h1(z,u)) -> h2#(s(x),y,h1(s(z),u))

and R consists of:

r1: f(j(x,y),y) -> g(f(x,k(y)))
r2: f(x,h1(y,z)) -> h2(|0|(),x,h1(y,z))
r3: g(h2(x,y,h1(z,u))) -> h2(s(x),y,h1(z,u))
r4: h2(x,j(y,h1(z,u)),h1(z,u)) -> h2(s(x),y,h1(s(z),u))
r5: i(f(x,h(y))) -> y
r6: i(h2(s(x),y,h1(x,z))) -> z
r7: k(h(x)) -> h1(|0|(),x)
r8: k(h1(x,y)) -> h1(s(x),y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        h2#_A(x1,x2,x3) = x1 + ((0,1),(1,1)) x2
        j_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,0)) x2
        h1_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        s_A(x1) = ((0,1),(0,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        h2#_A(x1,x2,x3) = (0,0)
        j_A(x1,x2) = ((0,1),(0,0)) x2 + (0,1)
        h1_A(x1,x2) = x2 + (1,1)
        s_A(x1) = (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.