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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = ((1,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = ((1,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        s#_A(x1) = ((0,1),(1,1)) x1
        +_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,0)
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        s#_A(x1) = (0,0)
        +_A(x1,x2) = (1,1)
        |0|_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        s#_A(x1) = (0,0)
        +_A(x1,x2) = (1,1)
        |0|_A() = (1,1)
    

The next rules are strictly ordered:

  p1

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