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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x1 + 4
          app_A(x1,x2) = x1 + x2 + 2
          uncurry_A() = 3
  
    precedence:
    
      app# = app = uncurry
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = [1]
      pi(uncurry) = []

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(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}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x1 + 2
          app_A(x1,x2) = x1 + x2 + 2
          uncurry_A() = 1
  
    precedence:
    
      app# = app = uncurry
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1]
      pi(uncurry) = []

The next rules are strictly ordered:

  p1

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