YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: +#(x,s(y)) -> s#(+(x,y))
p2: +#(x,s(y)) -> +#(x,y)
p3: s#(+(|0|(),y)) -> s#(y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,s(y)) -> s(+(x,y))
r3: +(|0|(),s(y)) -> s(y)
r4: s(+(|0|(),y)) -> s(y)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3}


-- Reduction pair.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,s(y)) -> s(+(x,y))
r3: +(|0|(),s(y)) -> s(y)
r4: s(+(|0|(),y)) -> 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(+#) = [2]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = [2]
        pi(s) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = 2
        pi(s) = 1
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,s(y)) -> s(+(x,y))
r3: +(|0|(),s(y)) -> s(y)
r4: s(+(|0|(),y)) -> s(y)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s# > |0| > +
      
      argument filter:
    
        pi(s#) = 1
        pi(+) = [1, 2]
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s# > |0| > +
      
      argument filter:
    
        pi(s#) = 1
        pi(+) = [1, 2]
        pi(|0|) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s# > |0| > +
      
      argument filter:
    
        pi(s#) = 1
        pi(+) = [1, 2]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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