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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        if#_A(x1,x2,x3) = max{27, x1 - 10, x2 - 4, x3 + 10}
        false_A = 56
        if2#_A(x1,x2) = max{36, x1 - 5, x2 + 9}
        weight#_A(x1) = max{27, x1 + 10}
        sum_A(x1,x2) = x2 + 3
        cons_A(x1,x2) = max{37, x2 + 11}
        |0|_A = 61
        tail_A(x1) = max{11, x1 - 10}
        empty_A(x1) = max{28, x1 + 20}
        s_A(x1) = 2
        nil_A = 10
        true_A = 29
  
    precedence:
    
      if# = false = if2# = weight# = sum = cons = |0| = tail = empty = s = nil = true
    
    partial status:
  
      pi(if#) = []
      pi(false) = []
      pi(if2#) = [2]
      pi(weight#) = [1]
      pi(sum) = []
      pi(cons) = [2]
      pi(|0|) = []
      pi(tail) = []
      pi(empty) = []
      pi(s) = []
      pi(nil) = []
      pi(true) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        sum#_A(x1,x2) = max{24, x1 + 6, x2 + 10}
        cons_A(x1,x2) = max{x1 + 1, x2 + 14}
        s_A(x1) = max{8, x1}
        |0|_A = 0
  
    precedence:
    
      sum# = cons = s = |0|
    
    partial status:
  
      pi(sum#) = [1, 2]
      pi(cons) = []
      pi(s) = [1]
      pi(|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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        sum#_A(x1,x2) = max{0, x1 - 6}
        cons_A(x1,x2) = max{8, x1 + 4}
        s_A(x1) = max{9, x1 + 5}
  
    precedence:
    
      s > sum# = cons
    
    partial status:
  
      pi(sum#) = []
      pi(cons) = [1]
      pi(s) = []

The next rules are strictly ordered:

  p1

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