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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          +#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,0)
          s_A(x1) = ((1,1),(1,1)) x1 + (1,3)
          +_A(x1,x2) = ((1,1),(1,1)) x1 + x2
          |0|_A() = (2,1)
  
    precedence:
    
      +# = s = + = |0|
    
    partial status:
  
      pi(+#) = []
      pi(s) = []
      pi(+) = [2]
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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