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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + (6,1)
          app_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,0)
          curry_A() = (3,0)
          plus_A() = (1,1)
          |0|_A() = (1,1)
          s_A() = (5,0)
  
    precedence:
    
      app# = app = curry = s > |0| > plus
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(curry) = []
      pi(plus) = []
      pi(|0|) = []
      pi(s) = []

The next rules are strictly ordered:

  p2

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#(f,x)

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#(f,x)

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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + (1,2)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,1)
          curry_A() = (2,1)
  
    precedence:
    
      app# = app = curry
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(curry) = []

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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (0,6)
          app_A(x1,x2) = x2 + (1,2)
          plus_A() = (3,5)
          s_A() = (4,1)
  
    precedence:
    
      app# = app > plus > s
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(plus) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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