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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        +#_A(x1,x2) = max{1, x1, x2}
        s_A(x1) = x1 + 1
  
    precedence:
    
      +# = s
    
    partial status:
  
      pi(+#) = [1, 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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        s#_A(x1) = max{5, x1 + 2}
        +_A(x1,x2) = x2 + 3
        |0|_A = 5
  
    precedence:
    
      s# = + = |0|
    
    partial status:
  
      pi(s#) = [1]
      pi(+) = [2]
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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