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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > |0| > plus > app# > app > curry
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(curry) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > |0| > plus > app > app# > curry
      
      argument filter:
    
        pi(app#) = 1
        pi(app) = [1, 2]
        pi(curry) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > |0| > plus > app > app# > curry
      
      argument filter:
    
        pi(app#) = []
        pi(app) = 1
        pi(curry) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
    

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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        app > plus > app# > s
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(plus) = []
        pi(s) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        plus > app > app# > s
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(plus) = []
        pi(s) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        plus > app > app# > s
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(plus) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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