YES

We show the termination of the TRS R:

  app(app(app(uncurry(),f),x),y) -> app(app(f,x),y)

-- SCC decomposition.

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

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

and R consists of:

r1: app(app(app(uncurry(),f),x),y) -> app(app(f,x),y)

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

and R consists of:

r1: app(app(app(uncurry(),f),x),y) -> app(app(f,x),y)

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 5, x2 + 2}
          app_A(x1,x2) = max{3, x1, x2 + 1}
          uncurry_A = 4
    
      precedence:
      
        app# = app = uncurry
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(uncurry) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 3, x2 + 3}
          app_A(x1,x2) = max{x1, x2 + 2}
          uncurry_A = 2
    
      precedence:
      
        app# = app = uncurry
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = [1, 2]
        pi(uncurry) = []
    

The next rules are strictly ordered:

  p1, p2

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