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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      +#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,1,1,1)) x2
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,1,0,0)) x1 + (2,1,1,2)
      +_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,1,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,0)) x2 + (0,3,0,0)
      |0|_A() = (1,1,1,1)

The next rules are strictly ordered:

  p1

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