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 path order with precedence:
  
    precedence:
    
      s > |0| > h1 > h > k > j > f#
    
    argument filter:
  
      pi(f#) = [1, 2]
      pi(j) = [1, 2]
      pi(k) = 1
      pi(h) = [1]
      pi(h1) = 2
      pi(|0|) = []
      pi(s) = 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 path order with precedence:
  
    precedence:
    
      s > h1 > j > h2#
    
    argument filter:
  
      pi(h2#) = [1, 2, 3]
      pi(j) = [1, 2]
      pi(h1) = 1
      pi(s) = 1

The next rules are strictly ordered:

  p1

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