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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1
        app_A(x1,x2) = x1 + x2
        uncurry_A() = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1
        app_A(x1,x2) = x1 + x2 + 1
        uncurry_A() = 0
    

The next rules are strictly ordered:

  p1, p2

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