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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = max{76, x1 - 48, x2 + 2, x3 + 23}
          false_A = 102
          if2#_A(x1,x2) = max{75, x1 + 1, x2 - 1}
          weight#_A(x1) = max{78, x1 + 24}
          sum_A(x1,x2) = max{45, x2 + 2}
          cons_A(x1,x2) = x2 + 72
          |0|_A = 40
          tail_A(x1) = max{4, x1 - 50}
          empty_A(x1) = x1 + 71
          s_A(x1) = max{43, x1 + 35}
          nil_A = 2
          true_A = 2
    
      precedence:
      
        cons = |0| = tail = s = nil = true > empty > if# = false > if2# > weight# = sum
      
      partial status:
    
        pi(if#) = []
        pi(false) = []
        pi(if2#) = []
        pi(weight#) = []
        pi(sum) = [2]
        pi(cons) = []
        pi(|0|) = []
        pi(tail) = []
        pi(empty) = [1]
        pi(s) = []
        pi(nil) = []
        pi(true) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = 44
          false_A = 14
          if2#_A(x1,x2) = 44
          weight#_A(x1) = 42
          sum_A(x1,x2) = max{41, x2}
          cons_A(x1,x2) = 42
          |0|_A = 15
          tail_A(x1) = 6
          empty_A(x1) = max{12, x1 - 27}
          s_A(x1) = 29
          nil_A = 7
          true_A = 13
    
      precedence:
      
        if# > empty = true > false > if2# = weight# = sum = cons = |0| = tail = s = nil
      
      partial status:
    
        pi(if#) = []
        pi(false) = []
        pi(if2#) = []
        pi(weight#) = []
        pi(sum) = [2]
        pi(cons) = []
        pi(|0|) = []
        pi(tail) = []
        pi(empty) = []
        pi(s) = []
        pi(nil) = []
        pi(true) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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)))))

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1,x2) = max{8, x1 + 4, x2 + 2}
          cons_A(x1,x2) = max{x1 + 2, x2 + 4}
          s_A(x1) = x1
          |0|_A = 2
    
      precedence:
      
        sum# = cons = s = |0|
      
      partial status:
    
        pi(sum#) = [1, 2]
        pi(cons) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1,x2) = x1 + 6
          cons_A(x1,x2) = max{x1 + 2, x2}
          s_A(x1) = x1 + 3
          |0|_A = 0
    
      precedence:
      
        sum# = cons = s = |0|
      
      partial status:
    
        pi(sum#) = [1]
        pi(cons) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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