YES
We show the termination of the TRS R:

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

Instead we prove termination of the uncurried system:

  app(uncurry_0(),y) -> uncurry_1(y)
  app(uncurry_1(x1),y) -> uncurry_2(x1,y)
  app(uncurry_2(x1,x2),y) -> uncurry_3(x1,x2,y)
  uncurry_3(f,x,y) -> app(app(f,x),y)


-- SCC decomposition.

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

p1: app#(uncurry_2(x1,x2),y) -> uncurry_3#(x1,x2,y)
p2: uncurry_3#(f,x,y) -> app#(app(f,x),y)
p3: uncurry_3#(f,x,y) -> app#(f,x)

and R consists of:

r1: app(uncurry_0(),y) -> uncurry_1(y)
r2: app(uncurry_1(x1),y) -> uncurry_2(x1,y)
r3: app(uncurry_2(x1,x2),y) -> uncurry_3(x1,x2,y)
r4: uncurry_3(f,x,y) -> app(app(f,x),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: app#(uncurry_2(x1,x2),y) -> uncurry_3#(x1,x2,y)
p2: uncurry_3#(f,x,y) -> app#(f,x)
p3: uncurry_3#(f,x,y) -> app#(app(f,x),y)

and R consists of:

r1: app(uncurry_0(),y) -> uncurry_1(y)
r2: app(uncurry_1(x1),y) -> uncurry_2(x1,y)
r3: app(uncurry_2(x1,x2),y) -> uncurry_3(x1,x2,y)
r4: uncurry_3(f,x,y) -> app(app(f,x),y)

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x1
          uncurry_2_A(x1,x2) = x1 + x2 + 4
          uncurry_3#_A(x1,x2,x3) = x1 + x2 + 3
          app_A(x1,x2) = x1 + x2 + 2
          uncurry_0_A() = 2
          uncurry_1_A(x1) = x1 + 3
          uncurry_3_A(x1,x2,x3) = x1 + x2 + x3 + 5
  
    base order 2:
  
      precedence:
      
        app# > uncurry_3# = uncurry_3 > app > uncurry_2 > uncurry_1 > uncurry_0
      
  
    partial status:
  
      pi(app#) = []
      pi(uncurry_2) = [1]
      pi(uncurry_3#) = [2]
      pi(app) = [2]
      pi(uncurry_0) = []
      pi(uncurry_1) = [1]
      pi(uncurry_3) = [2, 3]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: uncurry_3#(f,x,y) -> app#(f,x)
p2: uncurry_3#(f,x,y) -> app#(app(f,x),y)

and R consists of:

r1: app(uncurry_0(),y) -> uncurry_1(y)
r2: app(uncurry_1(x1),y) -> uncurry_2(x1,y)
r3: app(uncurry_2(x1,x2),y) -> uncurry_3(x1,x2,y)
r4: uncurry_3(f,x,y) -> app(app(f,x),y)

The estimated dependency graph contains the following SCCs:

  (no SCCs)