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:
  
      max/plus interpretation on natural numbers:
      
        if#_A(x1,x2,x3) = max{8, x1 - 35, x2 + 2, x3 - 29}
        false_A = 66
        if2#_A(x1,x2) = max{30, x1 - 1, x2 - 30}
        weight#_A(x1) = max{8, x1 - 28}
        sum_A(x1,x2) = max{27, x2 + 3}
        cons_A(x1,x2) = max{x1 + 83, x2 + 17}
        |0|_A = 1
        tail_A(x1) = max{7, x1 - 15}
        empty_A(x1) = max{6, x1 - 16}
        s_A(x1) = max{0, x1 - 57}
        nil_A = 6
        true_A = 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if#_A(x1,x2,x3) = 1
          false_A = 1
          if2#_A(x1,x2) = 1
          weight#_A(x1) = 1
          sum_A(x1,x2) = max{22, x2 + 1}
          cons_A(x1,x2) = max{0, x1 - 19, x2 - 19}
          |0|_A = 5
          tail_A(x1) = x1 + 19
          empty_A(x1) = max{1, x1 - 9}
          s_A(x1) = max{19, x1}
          nil_A = 0
          true_A = 1
          if#!_A(x1,x2,x3) = max{x1 + 10, x2 + 1, x3 + 11}
          false!_A = 3
          if2#!_A(x1,x2) = max{10, x1 + 1}
          weight#!_A(x1) = x1 + 12
          sum!_A(x1,x2) = 4
          cons!_A(x1,x2) = max{3, x1 - 21, x2 - 56}
          |0|!_A = 5
          tail!_A(x1) = max{13, x1 - 37}
          empty!_A(x1) = max{3, x1 - 58}
          s!_A(x1) = max{0, x1 - 1}
          nil!_A = 12
          true!_A = 1
  
    partial status:
  
      pi(if#) = []
      pi(false) = []
      pi(if2#) = []
      pi(weight#) = []
      pi(sum) = []
      pi(cons) = []
      pi(|0|) = []
      pi(tail) = []
      pi(empty) = []
      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:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1,x2) = 1
        cons_A(x1,x2) = max{5, x1 - 29, x2 - 29}
        s_A(x1) = max{0, x1 - 58}
        |0|_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1,x2) = max{0, x1 - 75, x2 - 37}
          cons_A(x1,x2) = max{x1 - 1, x2 + 36}
          s_A(x1) = x1
          |0|_A = 7
          sum#!_A(x1,x2) = max{4, x1 - 1}
          cons!_A(x1,x2) = max{3, x1 - 52, x2 - 53}
          s!_A(x1) = max{2, x1 - 53}
          |0|!_A = 1
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = []
      pi(s) = []
      pi(|0|) = []

The following rules are removed:

  p2



-- 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

  r3, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1,x2) = max{4, x1 + 1, x2 - 6}
        cons_A(x1,x2) = max{x1 + 13, x2 + 19}
        s_A(x1) = max{3, x1}
        sum_A(x1,x2) = max{22, x1 - 21, x2}
        nil_A = 0
        empty_A(x1) = max{21, x1 - 12}
        false_A = 20
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1,x2) = max{0, x1 - 26, x2 - 25}
          cons_A(x1,x2) = max{x1 - 9, x2 + 1}
          s_A(x1) = max{33, x1}
          sum_A(x1,x2) = max{76, x1 + 3, x2 + 52}
          nil_A = 24
          empty_A(x1) = max{89, x1 + 9}
          false_A = 1
          sum#!_A(x1,x2) = max{x1 + 48, x2 + 11}
          cons!_A(x1,x2) = max{x1 + 34, x2 + 49}
          s!_A(x1) = max{71, x1 - 69}
          sum!_A(x1,x2) = max{71, x1 - 25, x2 - 59}
          nil!_A = 0
          empty!_A(x1) = max{70, x1 - 81}
          false!_A = 69
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [1]
      pi(s) = [1]
      pi(sum) = [2]
      pi(nil) = []
      pi(empty) = []
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.