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:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = 8
        curry_2_A(x1,x2) = max{x1 - 6, x2 + 7}
        curry_3#_A(x1,x2,x3) = 8
        app_A(x1,x2) = max{x1 - 6, x2 + 7}
        curry_0_A = 1
        curry_1_A(x1) = max{2, x1}
        curry_3_A(x1,x2,x3) = max{x1 - 12, x2 + 1, x3 + 7}
        plus_0_A = 4
        plus_1_A(x1) = max{5, x1 - 1}
        plus_2_A(x1,x2) = max{3, x1 - 8, x2 + 2}
        s_0_A = 2
        s_1_A(x1) = max{3, x1 - 8}
        |0|_0_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 14, x2 - 8}
          curry_2_A(x1,x2) = max{x1 + 4, x2 + 13}
          curry_3#_A(x1,x2,x3) = max{0, x1 - 14, x2 - 1, x3 - 8}
          app_A(x1,x2) = max{x1 - 9, x2 + 13}
          curry_0_A = 10
          curry_1_A(x1) = x1 + 13
          curry_3_A(x1,x2,x3) = max{x1 - 5, x2 + 4, x3 + 13}
          plus_0_A = 0
          plus_1_A(x1) = max{8, x1 - 5}
          plus_2_A(x1,x2) = max{8, x1 - 14, x2 + 7}
          s_0_A = 4
          s_1_A(x1) = max{8, x1 - 15}
          |0|_0_A = 0
          app#!_A(x1,x2) = max{1, x1 - 1}
          curry_2!_A(x1,x2) = max{x1 + 8, x2 - 11}
          curry_3#!_A(x1,x2,x3) = max{x1 + 2, x2 + 12}
          app!_A(x1,x2) = max{8, x1 - 5, x2 - 10}
          curry_0!_A = 7
          curry_1!_A(x1) = max{6, x1 - 1}
          curry_3!_A(x1,x2,x3) = max{x1 - 14, x2 + 8, x3 - 10}
          plus_0!_A = 1
          plus_1!_A(x1) = max{2, x1 - 11}
          plus_2!_A(x1,x2) = max{10, x1 - 9, x2 - 3}
          s_0!_A = 1
          s_1!_A(x1) = max{9, x1 - 10}
          |0|_0!_A = 0
  
    partial status:
  
      pi(app#) = []
      pi(curry_2) = []
      pi(curry_3#) = []
      pi(app) = []
      pi(curry_0) = []
      pi(curry_1) = [1]
      pi(curry_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:

  p2



-- 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: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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#(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:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{1, x1}
        curry_2_A(x1,x2) = max{x1 + 8, x2 + 16}
        curry_3#_A(x1,x2,x3) = max{x1 + 5, x2 + 16}
        app_A(x1,x2) = max{x1 - 7, x2 + 16}
        curry_0_A = 1
        curry_1_A(x1) = x1 + 15
        curry_3_A(x1,x2,x3) = max{x1 + 1, x2 + 9, x3 + 16}
        plus_0_A = 25
        plus_1_A(x1) = max{17, x1 - 1}
        plus_2_A(x1,x2) = max{3, x1 - 9, x2 + 1}
        s_0_A = 10
        s_1_A(x1) = max{2, x1 - 4}
        |0|_0_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{1, x1 - 12, x2 - 2}
          curry_2_A(x1,x2) = max{11, x2 + 7}
          curry_3#_A(x1,x2,x3) = max{1, x2 - 5, x3 - 2}
          app_A(x1,x2) = max{11, x2 + 7}
          curry_0_A = 6
          curry_1_A(x1) = max{11, x1 + 7}
          curry_3_A(x1,x2,x3) = max{11, x3 + 7}
          plus_0_A = 4
          plus_1_A(x1) = 8
          plus_2_A(x1,x2) = max{8, x2 + 1}
          s_0_A = 4
          s_1_A(x1) = max{8, x1 - 7}
          |0|_0_A = 0
          app#!_A(x1,x2) = 1
          curry_2!_A(x1,x2) = max{7, x1 + 3}
          curry_3#!_A(x1,x2,x3) = max{6, x1 + 2}
          app!_A(x1,x2) = max{5, x1 - 3}
          curry_0!_A = 1
          curry_1!_A(x1) = 2
          curry_3!_A(x1,x2,x3) = max{8, x2 + 4}
          plus_0!_A = 1
          plus_1!_A(x1) = max{6, x1 - 10}
          plus_2!_A(x1,x2) = max{0, x1 - 9, x2 - 1}
          s_0!_A = 1
          s_1!_A(x1) = 0
          |0|_0!_A = 0
  
    partial status:
  
      pi(app#) = [1]
      pi(curry_2) = []
      pi(curry_3#) = []
      pi(app) = []
      pi(curry_0) = []
      pi(curry_1) = [1]
      pi(curry_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

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:
  
      max/plus interpretation on natural numbers:
      
        plus_2#_A(x1,x2) = x1 + 5
        s_1_A(x1) = x1 + 6
  
    base order 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{3, x1 - 3}
          plus_2#!_A(x1,x2) = max{0, x1 - 4}
          s_1!_A(x1) = max{0, x1 - 3}
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.