YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: -(|0|(),y) -> |0|()
r4: -(x,|0|()) -> x
r5: -(s(x),s(y)) -> -(x,y)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}


-- Reduction pair.

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

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

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: -(|0|(),y) -> |0|()
r4: -(x,|0|()) -> x
r5: -(s(x),s(y)) -> -(x,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(+#) = [1, 2]
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = 1
        pi(s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = [1]
        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(x),s(y)) -> -#(x,y)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: -(|0|(),y) -> |0|()
r4: -(x,|0|()) -> x
r5: -(s(x),s(y)) -> -(x,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(-#) = 1
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > -#
      
      argument filter:
    
        pi(-#) = 1
        pi(s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > -#
      
      argument filter:
    
        pi(-#) = 1
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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