YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: +#(s(x),s(y)) -> +#(s(x),+(y,|0|()))
p2: +#(s(x),s(y)) -> +#(y,|0|())

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: +#(s(x),s(y)) -> +#(s(x),+(y,|0|()))

and R consists of:

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

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 2
        +_A(x1,x2) = x1 + x2
        |0|_A() = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
        +_A(x1,x2) = x1 + 2
        |0|_A() = 1
    

The next rules are strictly ordered:

  p1

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