YES

We show the termination of the TRS R:

  sum(|0|()) -> |0|()
  sum(s(x)) -> +(sum(x),s(x))
  +(x,|0|()) -> x
  +(x,s(y)) -> s(+(x,y))

-- SCC decomposition.

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

p1: sum#(s(x)) -> +#(sum(x),s(x))
p2: sum#(s(x)) -> sum#(x)
p3: +#(x,s(y)) -> +#(x,y)

and R consists of:

r1: sum(|0|()) -> |0|()
r2: sum(s(x)) -> +(sum(x),s(x))
r3: +(x,|0|()) -> x
r4: +(x,s(y)) -> s(+(x,y))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3}


-- Reduction pair.

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

p1: sum#(s(x)) -> sum#(x)

and R consists of:

r1: sum(|0|()) -> |0|()
r2: sum(s(x)) -> +(sum(x),s(x))
r3: +(x,|0|()) -> x
r4: +(x,s(y)) -> s(+(x,y))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        sum#_A(x1) = x1 + 1
        s_A(x1) = x1
  
    precedence:
    
      sum# = s
    
    partial status:
  
      pi(sum#) = [1]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: +#(x,s(y)) -> +#(x,y)

and R consists of:

r1: sum(|0|()) -> |0|()
r2: sum(s(x)) -> +(sum(x),s(x))
r3: +(x,|0|()) -> x
r4: +(x,s(y)) -> s(+(x,y))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        +#_A(x1,x2) = max{1, x1, x2}
        s_A(x1) = x1 + 1
  
    precedence:
    
      +# = s
    
    partial status:
  
      pi(+#) = [1, 2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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