YES

We show the termination of the TRS R:

  app(nil(),k) -> k
  app(l,nil()) -> l
  app(cons(x,l),k) -> cons(x,app(l,k))
  sum(cons(x,nil())) -> cons(x,nil())
  sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
  a(h(),h(),x) -> s(x)
  a(x,s(y),h()) -> a(x,y,s(h()))
  a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
  a(s(x),h(),z) -> a(x,z,z)

-- SCC decomposition.

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

p1: app#(cons(x,l),k) -> app#(l,k)
p2: sum#(cons(x,cons(y,l))) -> sum#(cons(a(x,y,h()),l))
p3: sum#(cons(x,cons(y,l))) -> a#(x,y,h())
p4: a#(x,s(y),h()) -> a#(x,y,s(h()))
p5: a#(x,s(y),s(z)) -> a#(x,y,a(x,s(y),z))
p6: a#(x,s(y),s(z)) -> a#(x,s(y),z)
p7: a#(s(x),h(),z) -> a#(x,z,z)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}
  {p4, p5, p6, p7}


-- Reduction pair.

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

p1: app#(cons(x,l),k) -> app#(l,k)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{0, x1 - 2}
        cons_A(x1,x2) = max{x1 + 3, x2 + 1}
  
    precedence:
    
      app# = cons
    
    partial status:
  
      pi(app#) = []
      pi(cons) = [2]

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: sum#(cons(x,cons(y,l))) -> sum#(cons(a(x,y,h()),l))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The set of usable rules consists of

  r6, r7, r8, r9

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        sum#_A(x1) = max{41, x1 + 30}
        cons_A(x1,x2) = max{x1 + 17, x2 + 20}
        a_A(x1,x2,x3) = 14
        h_A = 16
        s_A(x1) = 14
  
    precedence:
    
      sum# = cons = a = h = s
    
    partial status:
  
      pi(sum#) = []
      pi(cons) = [2]
      pi(a) = []
      pi(h) = []
      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: a#(x,s(y),h()) -> a#(x,y,s(h()))
p2: a#(s(x),h(),z) -> a#(x,z,z)
p3: a#(x,s(y),s(z)) -> a#(x,s(y),z)
p4: a#(x,s(y),s(z)) -> a#(x,y,a(x,s(y),z))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The set of usable rules consists of

  r6, r7, r8, r9

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3) = max{x1 + 8, x2 + 2, x3 + 2}
        s_A(x1) = max{1, x1}
        h_A = 5
        a_A(x1,x2,x3) = max{x1 + 6, x2, x3}
  
    precedence:
    
      h > a > a# = s
    
    partial status:
  
      pi(a#) = [1]
      pi(s) = [1]
      pi(h) = []
      pi(a) = [1, 2, 3]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(x,s(y),h()) -> a#(x,y,s(h()))
p2: a#(x,s(y),s(z)) -> a#(x,s(y),z)
p3: a#(x,s(y),s(z)) -> a#(x,y,a(x,s(y),z))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: a#(x,s(y),h()) -> a#(x,y,s(h()))
p2: a#(x,s(y),s(z)) -> a#(x,y,a(x,s(y),z))
p3: a#(x,s(y),s(z)) -> a#(x,s(y),z)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The set of usable rules consists of

  r6, r7, r8, r9

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3) = max{x1 + 7, x2 + 8, x3 + 4}
        s_A(x1) = max{2, x1}
        h_A = 0
        a_A(x1,x2,x3) = max{x1 + 3, x2, x3}
  
    precedence:
    
      a > s = h > a#
    
    partial status:
  
      pi(a#) = [2]
      pi(s) = [1]
      pi(h) = []
      pi(a) = [1, 2, 3]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(x,s(y),s(z)) -> a#(x,y,a(x,s(y),z))
p2: a#(x,s(y),s(z)) -> a#(x,s(y),z)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: a#(x,s(y),s(z)) -> a#(x,y,a(x,s(y),z))
p2: a#(x,s(y),s(z)) -> a#(x,s(y),z)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The set of usable rules consists of

  r6, r7, r8, r9

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3) = max{x1 + 7, x2 + 7, x3 + 4}
        s_A(x1) = max{2, x1}
        a_A(x1,x2,x3) = max{x1 + 3, x2, x3}
        h_A = 1
  
    precedence:
    
      a > a# = s = h
    
    partial status:
  
      pi(a#) = [1, 2]
      pi(s) = [1]
      pi(a) = [1, 2, 3]
      pi(h) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(x,s(y),s(z)) -> a#(x,s(y),z)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: a#(x,s(y),s(z)) -> a#(x,s(y),z)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h()),l))
r6: a(h(),h(),x) -> s(x)
r7: a(x,s(y),h()) -> a(x,y,s(h()))
r8: a(x,s(y),s(z)) -> a(x,y,a(x,s(y),z))
r9: a(s(x),h(),z) -> a(x,z,z)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3) = max{0, x3 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      a# = s
    
    partial status:
  
      pi(a#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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