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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{9, x1 + 6, x2 + 3}
          s_A(x1) = x1 + 3
          +_A(x1,x2) = max{x1 + 3, x2 + 1}
          |0|_A = 2
    
      precedence:
      
        +# = s > + = |0|
      
      partial status:
    
        pi(+#) = [1, 2]
        pi(s) = [1]
        pi(+) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{x1 + 2, x2}
          s_A(x1) = max{4, x1 + 2}
          +_A(x1,x2) = max{x1, x2 + 2}
          |0|_A = 1
    
      precedence:
      
        +# = s = + = |0|
      
      partial status:
    
        pi(+#) = [2]
        pi(s) = []
        pi(+) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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