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 path order with precedence:
  
    precedence:
    
      app > id > |0| > plus > app# > s
    
    argument filter:
  
      pi(app#) = 1
      pi(app) = [1, 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.