YES

We show the termination of the TRS R:

  a(h(),h(),h(),x) -> s(x)
  a(l,x,s(y),h()) -> a(l,x,y,s(h()))
  a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
  a(l,s(x),h(),z) -> a(l,x,z,z)
  a(s(l),h(),h(),z) -> a(l,z,h(),z)
  +(x,h()) -> x
  +(h(),x) -> x
  +(s(x),s(y)) -> s(s(+(x,y)))
  +(+(x,y),z) -> +(x,+(y,z))
  s(h()) -> |1|()
  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(),h()),l))

-- SCC decomposition.

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

p1: a#(h(),h(),h(),x) -> s#(x)
p2: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p3: a#(l,x,s(y),h()) -> s#(h())
p4: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))
p5: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p6: a#(l,s(x),h(),z) -> a#(l,x,z,z)
p7: a#(s(l),h(),h(),z) -> a#(l,z,h(),z)
p8: +#(s(x),s(y)) -> s#(s(+(x,y)))
p9: +#(s(x),s(y)) -> s#(+(x,y))
p10: +#(s(x),s(y)) -> +#(x,y)
p11: +#(+(x,y),z) -> +#(x,+(y,z))
p12: +#(+(x,y),z) -> +#(y,z)
p13: app#(cons(x,l),k) -> app#(l,k)
p14: sum#(cons(x,cons(y,l))) -> sum#(cons(a(x,y,h(),h()),l))
p15: sum#(cons(x,cons(y,l))) -> a#(x,y,h(),h())

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The estimated dependency graph contains the following SCCs:

  {p14}
  {p2, p4, p5, p6, p7}
  {p10, p11, p12}
  {p13}


-- 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(),h()),l))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        h > |1| > a > s > cons > sum#
      
      argument filter:
    
        pi(sum#) = [1]
        pi(cons) = [2]
        pi(a) = [1, 2, 3, 4]
        pi(h) = []
        pi(s) = [1]
        pi(|1|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        sum#_A(x1) = x1
        cons_A(x1,x2) = 1
        a_A(x1,x2,x3,x4) = x1 + x2 + x3 + x4
        h_A() = 1
        s_A(x1) = x1 + 4
        |1|_A() = 0
    

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#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(s(l),h(),h(),z) -> a#(l,z,h(),z)
p3: a#(l,s(x),h(),z) -> a#(l,x,z,z)
p4: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p5: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        a# > a > s > h > |1|
      
      argument filter:
    
        pi(a#) = [1, 2, 3, 4]
        pi(s) = [1]
        pi(h) = []
        pi(a) = [1, 2, 3, 4]
        pi(|1|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1,x2,x3,x4) = x1 + x3
        s_A(x1) = 2
        h_A() = 1
        a_A(x1,x2,x3,x4) = x4 + 3
        |1|_A() = 0
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        +# > + > s > |1| > h
      
      argument filter:
    
        pi(+#) = 1
        pi(s) = [1]
        pi(+) = [1, 2]
        pi(h) = []
        pi(|1|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1
        s_A(x1) = 2
        +_A(x1,x2) = 1
        h_A() = 1
        |1|_A() = 0
    

The next rules are strictly ordered:

  p1, p2, p3

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

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        cons > app#
      
      argument filter:
    
        pi(app#) = [1]
        pi(cons) = [1, 2]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1
        cons_A(x1,x2) = x1 + x2 + 1
    

The next rules are strictly ordered:

  p1

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