YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = 1
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = 1
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1, p2

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