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 - 36, x2 - 2, x3 - 104}
            false_A = 1
            if2#_A(x1,x2) = max{0, x1 - 57, x2 - 104}
            weight#_A(x1) = max{0, x1 - 104}
            sum_A(x1,x2) = max{103, x1 - 60, x2}
            cons_A(x1,x2) = max{x1 - 2, x2 + 58}
            |0|_A = 68
            tail_A(x1) = max{43, x1 - 35}
            empty_A(x1) = max{1, x1 - 68}
            s_A(x1) = max{1, x1}
            nil_A = 0
            true_A = 1
            if#!_A(x1,x2,x3) = max{5, x1 - 36, x2 - 2, x3 - 105}
            false!_A = 3
            if2#!_A(x1,x2) = max{5, x1 - 3, x2 - 106}
            weight#!_A(x1) = max{5, x1 - 104}
            sum!_A(x1,x2) = max{6, x1 - 59, x2 - 67}
            cons!_A(x1,x2) = max{4, x1 - 123, x2 - 37}
            |0|!_A = 2
            tail!_A(x1) = max{34, x1 - 106}
            empty!_A(x1) = max{5, x1 - 69}
            s!_A(x1) = max{5, x1 - 70}
            nil!_A = 35
            true!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3) = max{0, x1 - 62, x2 - 30, x3 - 83}
            false_A = 40
            if2#_A(x1,x2) = max{0, x1 - 121, x2 - 83}
            weight#_A(x1) = max{0, x1 - 83}
            sum_A(x1,x2) = max{45, x1 - 63, x2}
            cons_A(x1,x2) = max{x1 - 2, x2 + 61}
            |0|_A = 45
            tail_A(x1) = max{21, x1 - 32}
            empty_A(x1) = max{0, x1 - 21}
            s_A(x1) = max{1, x1}
            nil_A = 20
            true_A = 0
            if#!_A(x1,x2,x3) = max{1, x1 - 35, x2 - 1, x3 - 45}
            false!_A = 2
            if2#!_A(x1,x2) = max{5, x1 - 1, x2 - 46}
            weight#!_A(x1) = max{4, x1 - 44}
            sum!_A(x1,x2) = max{42, x1 - 76, x2 - 83}
            cons!_A(x1,x2) = max{40, x1 - 87, x2 - 87}
            |0|!_A = 1
            tail!_A(x1) = max{39, x1 - 45}
            empty!_A(x1) = max{3, x1 - 17}
            s!_A(x1) = max{43, x1 - 79}
            nil!_A = 40
            true!_A = 2
        
  
    partial status:
  
      pi(if#) = []
      pi(false) = []
      pi(if2#) = []
      pi(weight#) = []
      pi(sum) = [1, 2]
      pi(cons) = [1, 2]
      pi(|0|) = []
      pi(tail) = [1]
      pi(empty) = []
      pi(s) = [1]
      pi(nil) = []
      pi(true) = []

The following rules are removed:

  p2, p3



-- SCC decomposition.

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

p1: if#(false(),b,x) -> if2#(b,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:

  (no SCCs)

-- 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

  r5

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 - 30, x2 - 29}
            cons_A(x1,x2) = max{x1 + 28, x2 + 28}
            s_A(x1) = x1
            |0|_A = 18
            empty_A(x1) = max{16, x1 - 28}
            false_A = 1
            sum#!_A(x1,x2) = max{24, x1 - 20}
            cons!_A(x1,x2) = max{24, x1 + 5, x2 - 19}
            s!_A(x1) = max{25, x1 - 17}
            |0|!_A = 3
            empty!_A(x1) = max{6, x1 - 29}
            false!_A = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1,x2) = max{0, x1 - 24, x2 - 23}
            cons_A(x1,x2) = max{x1 + 17, x2}
            s_A(x1) = max{5, x1}
            |0|_A = 27
            empty_A(x1) = max{13, x1 - 17}
            false_A = 1
            sum#!_A(x1,x2) = max{7, x1 - 37, x2 - 36}
            cons!_A(x1,x2) = max{13, x1 - 37, x2 - 37}
            s!_A(x1) = max{8, x1 - 27}
            |0|!_A = 8
            empty!_A(x1) = max{12, x1 - 31}
            false!_A = 10
        
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = [2]
      pi(s) = [1]
      pi(|0|) = []
      pi(empty) = [1]
      pi(false) = []

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

  r5

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 - 1}
            cons_A(x1,x2) = max{0, x1 - 6}
            s_A(x1) = max{19, x1 + 8}
            empty_A(x1) = max{25, x1 - 10}
            false_A = 1
            sum#!_A(x1,x2) = x1 + 2
            cons!_A(x1,x2) = max{x1 + 29, x2 + 23}
            s!_A(x1) = max{27, x1 - 5}
            empty!_A(x1) = max{0, x1 - 8}
            false!_A = 21
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1,x2) = max{x1 - 38, x2 + 1}
            cons_A(x1,x2) = max{1, x2 - 1}
            s_A(x1) = x1 + 27
            empty_A(x1) = max{1, x1 - 4}
            false_A = 1
            sum#!_A(x1,x2) = x1 + 32
            cons!_A(x1,x2) = x1 + 1
            s!_A(x1) = 29
            empty!_A(x1) = max{33, x1 - 11}
            false!_A = 32
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [1]
      pi(s) = [1]
      pi(empty) = [1]
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.