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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3) = max{0, x1 - 63, x2 - 26, x3 - 64}
            false_A = 56
            if2#_A(x1,x2) = max{0, x1 - 96, x2 - 64}
            weight#_A(x1) = max{0, x1 - 64}
            sum_A(x1,x2) = max{63, x1 - 58, x2}
            cons_A(x1,x2) = x2 + 57
            |0|_A = 54
            tail_A(x1) = max{5, x1 - 37}
            empty_A(x1) = max{4, x1 - 1}
            s_A(x1) = x1
            nil_A = 0
            true_A = 0
            if#!_A(x1,x2,x3) = max{x1 - 10, x2 + 30, x3 - 8}
            false!_A = 0
            if2#!_A(x1,x2) = max{30, x1, x2 - 9}
            weight#!_A(x1) = max{34, x1 - 7}
            sum!_A(x1,x2) = max{29, x1 - 10, x2 + 7}
            cons!_A(x1,x2) = max{66, x1 - 64, x2 + 39}
            |0|!_A = 69
            tail!_A(x1) = 29
            empty!_A(x1) = max{1, x1 - 58}
            s!_A(x1) = 65
            nil!_A = 21
            true!_A = 20
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if#_A(x1,x2,x3) = 0
              false_A() = 10
              if2#_A(x1,x2) = 0
              weight#_A(x1) = 0
              sum_A(x1,x2) = x2
              cons_A(x1,x2) = x2 + 5
              |0|_A() = 1
              tail_A(x1) = x1 + 1
              empty_A(x1) = 10
              s_A(x1) = x1 + 1
              nil_A() = 1
              true_A() = 10
              if#!_A(x1,x2,x3) = x3 + 1
              false!_A() = 5
              if2#!_A(x1,x2) = x1 + x2
              weight#!_A(x1) = x1 + 2
              sum!_A(x1,x2) = x1 + x2 + 2
              cons!_A(x1,x2) = 11
              |0|!_A() = 11
              tail!_A(x1) = x1 + 3
              empty!_A(x1) = 3
              s!_A(x1) = 10
              nil!_A() = 2
              true!_A() = 1
        
  
    partial status:
  
      pi(if#) = []
      pi(false) = []
      pi(if2#) = []
      pi(weight#) = []
      pi(sum) = [1, 2]
      pi(cons) = [2]
      pi(|0|) = []
      pi(tail) = [1]
      pi(empty) = [1]
      pi(s) = []
      pi(nil) = []
      pi(true) = []

The following rules are removed:

  p1, p2, p3

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1,x2) = max{0, x1 - 63, x2 - 29}
            cons_A(x1,x2) = max{x1 - 5, x2 + 28}
            s_A(x1) = x1
            |0|_A = 22
            sum#!_A(x1,x2) = max{35, x1 + 7, x2}
            cons!_A(x1,x2) = x2 + 5
            s!_A(x1) = 5
            |0|!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1,x2) = x1 + x2
              cons_A(x1,x2) = x1 + x2 + 2
              s_A(x1) = x1 + 1
              |0|_A() = 1
              sum#!_A(x1,x2) = x1 + 1
              cons!_A(x1,x2) = 2
              s!_A(x1) = x1 + 1
              |0|!_A() = 8
        
  
    partial status:
  
      pi(sum#) = [2]
      pi(cons) = []
      pi(s) = [1]
      pi(|0|) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.