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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((0,0),(1,0)) x1 + (1,1)
            app_A(x1,x2) = ((1,0),(0,0)) x2 + (3,8)
            plus_A() = (4,10)
            s_A() = (2,9)
            |0|_A() = (1,12)
            id_A() = (0,11)
    
      precedence:
      
        plus = s > |0| > app# = app = id
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(plus) = []
        pi(s) = []
        pi(|0|) = []
        pi(id) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (0,0)
            app_A(x1,x2) = (3,3)
            plus_A() = (1,1)
            s_A() = (2,2)
            |0|_A() = (3,3)
            id_A() = (2,2)
    
      precedence:
      
        id > plus > |0| > app > app# = s
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        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.