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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 9, x2 + 7}
          j_A(x1,x2) = max{x1 + 5, x2 + 5}
          k_A(x1) = max{5, x1 + 4}
          h_A(x1) = x1 + 9
          h1_A(x1,x2) = max{x1 + 2, x2 + 2}
          |0|_A = 6
          s_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        f# = j = h > h1 = |0| > k = s
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(j) = [1, 2]
        pi(k) = [1]
        pi(h) = [1]
        pi(h1) = [1, 2]
        pi(|0|) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 4, x2 + 3}
          j_A(x1,x2) = max{x1 - 1, x2}
          k_A(x1) = x1 + 5
          h_A(x1) = x1 + 4
          h1_A(x1,x2) = max{x1 + 11, x2 + 14}
          |0|_A = 3
          s_A(x1) = max{3, x1}
    
      precedence:
      
        f# = j = k = h = h1 = |0| = s
      
      partial status:
    
        pi(f#) = []
        pi(j) = [2]
        pi(k) = [1]
        pi(h) = [1]
        pi(h1) = [1, 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 combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          h2#_A(x1,x2,x3) = max{x1 + 1, x2 + 13, x3 + 4}
          j_A(x1,x2) = max{x1 + 3, x2}
          h1_A(x1,x2) = max{x1 + 3, x2 + 3}
          s_A(x1) = 2
    
      precedence:
      
        h2# = j = h1 = s
      
      partial status:
    
        pi(h2#) = [1]
        pi(j) = [2]
        pi(h1) = [1]
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          h2#_A(x1,x2,x3) = x1 + 4
          j_A(x1,x2) = x2 + 7
          h1_A(x1,x2) = max{12, x1 + 6}
          s_A(x1) = 5
    
      precedence:
      
        h2# = j = h1 = s
      
      partial status:
    
        pi(h2#) = []
        pi(j) = [2]
        pi(h1) = [1]
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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