YES

We show the termination of the TRS R:

  app(id(),x) -> x
  app(plus(),|0|()) -> id()
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

-- SCC decomposition.

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

p1: app#(app(plus(),app(s(),x)),y) -> app#(s(),app(app(plus(),x),y))
p2: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)
p3: app#(app(plus(),app(s(),x)),y) -> app#(plus(),x)

and R consists of:

r1: app(id(),x) -> x
r2: app(plus(),|0|()) -> id()
r3: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)

and R consists of:

r1: app(id(),x) -> x
r2: app(plus(),|0|()) -> id()
r3: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

The set of usable rules consists of

  r2

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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