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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{14, x1 + 7}
        app_A(x1,x2) = max{19, x1 - 9, x2 + 17}
        uncurry_A = 31
  
    precedence:
    
      app# = app = uncurry
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = [2]
      pi(uncurry) = []

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

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}


-- 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)

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
        app_A(x1,x2) = max{x1, x2 + 1}
        uncurry_A = 1
  
    precedence:
    
      app# = app = uncurry
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = [1, 2]
      pi(uncurry) = []

The next rules are strictly ordered:

  p1

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