YES

We show the termination of the TRS R:

  sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
  sum(cons(|0|(),x),y) -> sum(x,y)
  sum(nil(),y) -> y
  empty(nil()) -> true()
  empty(cons(n,x)) -> false()
  tail(nil()) -> nil()
  tail(cons(n,x)) -> x
  head(cons(n,x)) -> n
  weight(x) -> if(empty(x),empty(tail(x)),x)
  if(true(),b,x) -> weight_undefined_error()
  if(false(),b,x) -> if2(b,x)
  if2(true(),x) -> head(x)
  if2(false(),x) -> weight(sum(x,cons(|0|(),tail(tail(x)))))

-- SCC decomposition.

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

p1: sum#(cons(s(n),x),cons(m,y)) -> sum#(cons(n,x),cons(s(m),y))
p2: sum#(cons(|0|(),x),y) -> sum#(x,y)
p3: weight#(x) -> if#(empty(x),empty(tail(x)),x)
p4: weight#(x) -> empty#(x)
p5: weight#(x) -> empty#(tail(x))
p6: weight#(x) -> tail#(x)
p7: if#(false(),b,x) -> if2#(b,x)
p8: if2#(true(),x) -> head#(x)
p9: if2#(false(),x) -> weight#(sum(x,cons(|0|(),tail(tail(x)))))
p10: if2#(false(),x) -> sum#(x,cons(|0|(),tail(tail(x))))
p11: if2#(false(),x) -> tail#(tail(x))
p12: if2#(false(),x) -> tail#(x)

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: empty(nil()) -> true()
r5: empty(cons(n,x)) -> false()
r6: tail(nil()) -> nil()
r7: tail(cons(n,x)) -> x
r8: head(cons(n,x)) -> n
r9: weight(x) -> if(empty(x),empty(tail(x)),x)
r10: if(true(),b,x) -> weight_undefined_error()
r11: if(false(),b,x) -> if2(b,x)
r12: if2(true(),x) -> head(x)
r13: if2(false(),x) -> weight(sum(x,cons(|0|(),tail(tail(x)))))

The estimated dependency graph contains the following SCCs:

  {p3, p7, p9}
  {p1, p2}


-- Reduction pair.

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

p1: if#(false(),b,x) -> if2#(b,x)
p2: if2#(false(),x) -> weight#(sum(x,cons(|0|(),tail(tail(x)))))
p3: weight#(x) -> if#(empty(x),empty(tail(x)),x)

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: empty(nil()) -> true()
r5: empty(cons(n,x)) -> false()
r6: tail(nil()) -> nil()
r7: tail(cons(n,x)) -> x
r8: head(cons(n,x)) -> n
r9: weight(x) -> if(empty(x),empty(tail(x)),x)
r10: if(true(),b,x) -> weight_undefined_error()
r11: if(false(),b,x) -> if2(b,x)
r12: if2(true(),x) -> head(x)
r13: if2(false(),x) -> weight(sum(x,cons(|0|(),tail(tail(x)))))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      if#_A(x1,x2,x3) = max{1, x2 - 14, x3 - 13}
      false_A = 18
      if2#_A(x1,x2) = max{0, x1 - 15, x2 - 14}
      weight#_A(x1) = max{2, x1 - 13}
      sum_A(x1,x2) = max{12, x2 + 1}
      cons_A(x1,x2) = max{13, x1 + 10, x2 + 10}
      |0|_A = 3
      tail_A(x1) = max{1, x1 - 9}
      empty_A(x1) = x1 + 9
      s_A(x1) = 2
      nil_A = 0
      true_A = 1
    
    2. max/plus interpretations on natural numbers:
    
      if#_A(x1,x2,x3) = 0
      false_A = 1
      if2#_A(x1,x2) = 2
      weight#_A(x1) = 1
      sum_A(x1,x2) = 2
      cons_A(x1,x2) = 2
      |0|_A = 0
      tail_A(x1) = 3
      empty_A(x1) = x1
      s_A(x1) = 0
      nil_A = 2
      true_A = 1
    

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: sum#(cons(s(n),x),cons(m,y)) -> sum#(cons(n,x),cons(s(m),y))
p2: sum#(cons(|0|(),x),y) -> sum#(x,y)

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: empty(nil()) -> true()
r5: empty(cons(n,x)) -> false()
r6: tail(nil()) -> nil()
r7: tail(cons(n,x)) -> x
r8: head(cons(n,x)) -> n
r9: weight(x) -> if(empty(x),empty(tail(x)),x)
r10: if(true(),b,x) -> weight_undefined_error()
r11: if(false(),b,x) -> if2(b,x)
r12: if2(true(),x) -> head(x)
r13: if2(false(),x) -> weight(sum(x,cons(|0|(),tail(tail(x)))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      sum#_A(x1,x2) = x1 + 5
      cons_A(x1,x2) = max{x1 - 1, x2 + 1}
      s_A(x1) = max{1, x1}
      |0|_A = 0
    
    2. max/plus interpretations on natural numbers:
    
      sum#_A(x1,x2) = 0
      cons_A(x1,x2) = 0
      s_A(x1) = 0
      |0|_A = 0
    

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: sum#(cons(s(n),x),cons(m,y)) -> sum#(cons(n,x),cons(s(m),y))

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: empty(nil()) -> true()
r5: empty(cons(n,x)) -> false()
r6: tail(nil()) -> nil()
r7: tail(cons(n,x)) -> x
r8: head(cons(n,x)) -> n
r9: weight(x) -> if(empty(x),empty(tail(x)),x)
r10: if(true(),b,x) -> weight_undefined_error()
r11: if(false(),b,x) -> if2(b,x)
r12: if2(true(),x) -> head(x)
r13: if2(false(),x) -> weight(sum(x,cons(|0|(),tail(tail(x)))))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: sum#(cons(s(n),x),cons(m,y)) -> sum#(cons(n,x),cons(s(m),y))

and R consists of:

r1: sum(cons(s(n),x),cons(m,y)) -> sum(cons(n,x),cons(s(m),y))
r2: sum(cons(|0|(),x),y) -> sum(x,y)
r3: sum(nil(),y) -> y
r4: empty(nil()) -> true()
r5: empty(cons(n,x)) -> false()
r6: tail(nil()) -> nil()
r7: tail(cons(n,x)) -> x
r8: head(cons(n,x)) -> n
r9: weight(x) -> if(empty(x),empty(tail(x)),x)
r10: if(true(),b,x) -> weight_undefined_error()
r11: if(false(),b,x) -> if2(b,x)
r12: if2(true(),x) -> head(x)
r13: if2(false(),x) -> weight(sum(x,cons(|0|(),tail(tail(x)))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      sum#_A(x1,x2) = x1 + 1
      cons_A(x1,x2) = x1 + 1
      s_A(x1) = max{2, x1 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      sum#_A(x1,x2) = 0
      cons_A(x1,x2) = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1

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