YES
We show the termination of the TRS R:

  app(app(plus(),|0|()),y) -> y
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
  app(app(app(curry(),f),x),y) -> app(app(f,x),y)
  add() -> app(curry(),plus())

Instead we prove termination of the uncurried system:

  app(curry_0(),y) -> curry_1(y)
  app(curry_1(x1),y) -> curry_2(x1,y)
  app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
  app(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))
  curry_3(f,x,y) -> app(app(f,x),y)
  add_0() -> curry_1(plus_0())


-- SCC decomposition.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: app#(plus_1(x1),y) -> plus_2#(x1,y)
p3: plus_2#(s_1(x),y) -> plus_2#(x,y)
p4: curry_3#(f,x,y) -> app#(app(f,x),y)
p5: curry_3#(f,x,y) -> app#(f,x)

and R consists of:

r1: app(curry_0(),y) -> curry_1(y)
r2: app(curry_1(x1),y) -> curry_2(x1,y)
r3: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r4: app(plus_0(),y) -> plus_1(y)
r5: app(plus_1(x1),y) -> plus_2(x1,y)
r6: app(s_0(),y) -> s_1(y)
r7: plus_2(|0|_0(),y) -> y
r8: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r9: curry_3(f,x,y) -> app(app(f,x),y)
r10: add_0() -> curry_1(plus_0())

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5}
  {p3}


-- Reduction pair.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: curry_3#(f,x,y) -> app#(f,x)
p3: curry_3#(f,x,y) -> app#(app(f,x),y)

and R consists of:

r1: app(curry_0(),y) -> curry_1(y)
r2: app(curry_1(x1),y) -> curry_2(x1,y)
r3: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r4: app(plus_0(),y) -> plus_1(y)
r5: app(plus_1(x1),y) -> plus_2(x1,y)
r6: app(s_0(),y) -> s_1(y)
r7: plus_2(|0|_0(),y) -> y
r8: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r9: curry_3(f,x,y) -> app(app(f,x),y)
r10: add_0() -> curry_1(plus_0())

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

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:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              curry_2_A(x1,x2) = x1 + x2 + 3
              curry_3#_A(x1,x2,x3) = x1 + x2 + 1
              app_A(x1,x2) = x1 + x2 + 1
              curry_0_A() = 3
              curry_1_A(x1) = x1 + 4
              curry_3_A(x1,x2,x3) = x1 + x2 + x3 + 2
              plus_0_A() = 1
              plus_1_A(x1) = x1 + 1
              plus_2_A(x1,x2) = x1 + x2 + 1
              s_0_A() = 1
              s_1_A(x1) = 1
              |0|_0_A() = 1
              app#!_A(x1,x2) = x1
              curry_2!_A(x1,x2) = x1 + x2 + 6
              curry_3#!_A(x1,x2,x3) = x1 + x2 + 2
              app!_A(x1,x2) = x1 + x2 + 3
              curry_0!_A() = 5
              curry_1!_A(x1) = 5
              curry_3!_A(x1,x2,x3) = x1 + x2 + x3 + 5
              plus_0!_A() = 0
              plus_1!_A(x1) = 1
              plus_2!_A(x1,x2) = x2 + 2
              s_0!_A() = 1
              s_1!_A(x1) = 0
              |0|_0!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 1
            curry_2_A(x1,x2) = x2 + 6
            curry_3#_A(x1,x2,x3) = 1
            app_A(x1,x2) = x2 + 8
            curry_0_A = 1
            curry_1_A(x1) = max{1, x1 - 13}
            curry_3_A(x1,x2,x3) = x3 + 8
            plus_0_A = 0
            plus_1_A(x1) = max{7, x1 - 6}
            plus_2_A(x1,x2) = max{7, x2 + 1}
            s_0_A = 1
            s_1_A(x1) = max{7, x1 - 9}
            |0|_0_A = 0
            app#!_A(x1,x2) = max{x1 - 4, x2 + 3}
            curry_2!_A(x1,x2) = max{3, x1 - 14}
            curry_3#!_A(x1,x2,x3) = 1
            app!_A(x1,x2) = max{4, x1 - 2, x2 - 10}
            curry_0!_A = 1
            curry_1!_A(x1) = max{2, x1 - 11}
            curry_3!_A(x1,x2,x3) = max{0, x1 - 3, x2 - 11, x3 - 11}
            plus_0!_A = 1
            plus_1!_A(x1) = max{4, x1 - 10}
            plus_2!_A(x1,x2) = max{5, x1 - 8}
            s_0!_A = 1
            s_1!_A(x1) = max{5, x1 - 9}
            |0|_0!_A = 0
        
  
    partial status:
  
      pi(app#) = []
      pi(curry_2) = []
      pi(curry_3#) = []
      pi(app) = [2]
      pi(curry_0) = []
      pi(curry_1) = []
      pi(curry_3) = [3]
      pi(plus_0) = []
      pi(plus_1) = []
      pi(plus_2) = [2]
      pi(s_0) = []
      pi(s_1) = []
      pi(|0|_0) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- 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(curry_0(),y) -> curry_1(y)
r2: app(curry_1(x1),y) -> curry_2(x1,y)
r3: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r4: app(plus_0(),y) -> plus_1(y)
r5: app(plus_1(x1),y) -> plus_2(x1,y)
r6: app(s_0(),y) -> s_1(y)
r7: plus_2(|0|_0(),y) -> y
r8: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r9: curry_3(f,x,y) -> app(app(f,x),y)
r10: add_0() -> curry_1(plus_0())

The set of usable rules consists of

  (no rules)

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:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus_2#_A(x1,x2) = x1
              s_1_A(x1) = x1 + 2
              plus_2#!_A(x1,x2) = x1
              s_1!_A(x1) = x1 + 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{4, x1 - 4}
            plus_2#!_A(x1,x2) = 6
            s_1!_A(x1) = x1 + 7
        
  
    partial status:
  
      pi(plus_2#) = []
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.