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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,3)
            j_A(x1,x2) = x1 + (4,6)
            k_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            h_A(x1) = x1 + (1,1)
            h1_A(x1,x2) = x2
            |0|_A() = (0,0)
            s_A(x1) = (1,1)
    
      precedence:
      
        f# = j = k = h = h1 = |0| = s
      
      partial status:
    
        pi(f#) = []
        pi(j) = []
        pi(k) = []
        pi(h) = []
        pi(h1) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = (0,0)
            j_A(x1,x2) = ((1,0),(0,0)) x1 + (1,1)
            k_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            h_A(x1) = (3,2)
            h1_A(x1,x2) = (2,3)
            |0|_A() = (1,1)
            s_A(x1) = (1,1)
    
      precedence:
      
        j > k = h1 = s > f# = h = |0|
      
      partial status:
    
        pi(f#) = []
        pi(j) = []
        pi(k) = []
        pi(h) = []
        pi(h1) = []
        pi(|0|) = []
        pi(s) = []
    

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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            h2#_A(x1,x2,x3) = x2 + ((1,0),(1,1)) x3 + (1,1)
            j_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (3,7)
            h1_A(x1,x2) = x1 + (2,1)
            s_A(x1) = (1,2)
    
      precedence:
      
        j = h1 = s > h2#
      
      partial status:
    
        pi(h2#) = [2]
        pi(j) = [1, 2]
        pi(h1) = [1]
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            h2#_A(x1,x2,x3) = (0,0)
            j_A(x1,x2) = x1 + x2 + (1,1)
            h1_A(x1,x2) = (1,0)
            s_A(x1) = (1,0)
    
      precedence:
      
        j = s > h2# = h1
      
      partial status:
    
        pi(h2#) = []
        pi(j) = []
        pi(h1) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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