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

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            plus_2#_A(0,0) = 0
            plus_2#_A(0,1) = 0
            plus_2#_A(1,0) = 0
            plus_2#_A(1,1) = 0
            s_1_A(0) = 0
            s_1_A(1) = 1
            app_A(0,0) = 0
            app_A(0,1) = 1
            app_A(1,0) = 0
            app_A(1,1) = 0
            id_0_A = 0
            id_1_A(0) = 0
            id_1_A(1) = 1
            plus_0_A = 1
            plus_1_A(0) = 0
            plus_1_A(1) = 0
            plus_2_A(0,0) = 0
            plus_2_A(0,1) = 1
            plus_2_A(1,0) = 0
            plus_2_A(1,1) = 1
            s_0_A = 0
            |0|_0_A = 1
            plus_2#!_A(0,0) = 0
            plus_2#!_A(0,1) = 0
            plus_2#!_A(1,0) = 0
            plus_2#!_A(1,1) = 0
            s_1!_A(0) = 0
            s_1!_A(1) = 0
            app!_A(0,0) = 1
            app!_A(0,1) = 1
            app!_A(1,0) = 1
            app!_A(1,1) = 1
            id_0!_A = 0
            id_1!_A(0) = 0
            id_1!_A(1) = 0
            plus_0!_A = 0
            plus_1!_A(0) = 0
            plus_1!_A(1) = 0
            plus_2!_A(0,0) = 1
            plus_2!_A(0,1) = 1
            plus_2!_A(1,0) = 1
            plus_2!_A(1,1) = 1
            s_0!_A = 0
            |0|_0!_A = 0
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]
      pi(app) = [1, 2]
      pi(id_0) = []
      pi(id_1) = [1]
      pi(plus_0) = []
      pi(plus_1) = [1]
      pi(plus_2) = [1, 2]
      pi(s_0) = []
      pi(|0|_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.