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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{11, x1 + 7, x2 + 6}
        app_A(x1,x2) = max{x1 + 5, x2}
        curry_A = 0
        plus_A = 2
        |0|_A = 0
        s_A = 1
  
    precedence:
    
      app# = app > curry = plus = |0| = s
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(curry) = []
      pi(plus) = []
      pi(|0|) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{x1, x2 + 2}
        app_A(x1,x2) = max{2, x1, x2}
        curry_A = 2
        plus_A = 1
        |0|_A = 0
        s_A = 3
  
    precedence:
    
      app# = app = curry = plus = |0| = s
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = [1, 2]
      pi(curry) = []
      pi(plus) = []
      pi(|0|) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{0, x1 - 9}
        app_A(x1,x2) = max{x1 + 2, x2 + 4}
        plus_A = 2
        s_A = 6
  
    precedence:
    
      app# = app = plus = s
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 2]
      pi(plus) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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