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())

-- 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)
p4: app#(app(app(curry(),f),x),y) -> app#(app(f,x),y)
p5: app#(app(app(curry(),f),x),y) -> app#(f,x)
p6: add#() -> app#(curry(),plus())

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(app(curry(),f),x),y) -> app(app(f,x),y)
r4: add() -> app(curry(),plus())

The estimated dependency graph contains the following SCCs:

  {p4, p5}
  {p2}


-- Reduction pair.

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

p1: app#(app(app(curry(),f),x),y) -> app#(f,x)
p2: app#(app(app(curry(),f),x),y) -> app#(app(f,x),y)

and R consists of:

r1: app(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(app(curry(),f),x),y) -> app(app(f,x),y)
r4: add() -> app(curry(),plus())

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,0,1,1)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2
      app_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x2 + (1,1,1,0)
      curry_A() = (1,1,0,0)
      plus_A() = (1,0,0,0)
      |0|_A() = (1,1,0,0)
      s_A() = (0,0,1,0)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.

-- 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(app(plus(),|0|()),y) -> y
r2: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
r3: app(app(app(curry(),f),x),y) -> app(app(f,x),y)
r4: add() -> app(curry(),plus())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,1)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x2
      app_A(x1,x2) = x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,1,1)) x2 + (0,0,1,0)
      plus_A() = (1,1,1,1)
      s_A() = (0,0,1,1)

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.