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))

Instead we prove termination of the uncurried system:

  app(id_0(),y) -> id_1(y)
  app(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  id_1(x) -> x
  plus_1(|0|_0()) -> id_0()
  plus_2(|0|_0(),y1) -> id_1(y1)
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))


-- SCC decomposition.

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

p1: app#(id_0(),y) -> id_1#(y)
p2: app#(plus_0(),y) -> plus_1#(y)
p3: app#(plus_1(x1),y) -> plus_2#(x1,y)
p4: plus_2#(|0|_0(),y1) -> id_1#(y1)
p5: plus_2#(s_1(x),y) -> plus_2#(x,y)

and R consists of:

r1: app(id_0(),y) -> id_1(y)
r2: app(plus_0(),y) -> plus_1(y)
r3: app(plus_1(x1),y) -> plus_2(x1,y)
r4: app(s_0(),y) -> s_1(y)
r5: id_1(x) -> x
r6: plus_1(|0|_0()) -> id_0()
r7: plus_2(|0|_0(),y1) -> id_1(y1)
r8: plus_2(s_1(x),y) -> s_1(plus_2(x,y))

The estimated dependency graph contains the following SCCs:

  {p5}


-- Reduction pair.

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

p1: plus_2#(s_1(x),y) -> plus_2#(x,y)

and R consists of:

r1: app(id_0(),y) -> id_1(y)
r2: app(plus_0(),y) -> plus_1(y)
r3: app(plus_1(x1),y) -> plus_2(x1,y)
r4: app(s_0(),y) -> s_1(y)
r5: id_1(x) -> x
r6: plus_1(|0|_0()) -> id_0()
r7: plus_2(|0|_0(),y1) -> id_1(y1)
r8: plus_2(s_1(x),y) -> s_1(plus_2(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus_2#_A(x1,x2) = max{3, x1 + 1, x2 + 1}
        s_1_A(x1) = max{2, x1}
  
    base order 2:
  
      precedence:
      
        plus_2# > s_1
      
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.