YES
We show the termination of the TRS R:

  app(app(apply(),f),x) -> app(f,x)

Instead we prove termination of the uncurried system:

  app(apply_0(),y) -> apply_1(y)
  app(apply_1(x1),y) -> apply_2(x1,y)
  apply_2(f,x) -> app(f,x)


-- SCC decomposition.

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

p1: app#(apply_1(x1),y) -> apply_2#(x1,y)
p2: apply_2#(f,x) -> app#(f,x)

and R consists of:

r1: app(apply_0(),y) -> apply_1(y)
r2: app(apply_1(x1),y) -> apply_2(x1,y)
r3: apply_2(f,x) -> app(f,x)

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#(apply_1(x1),y) -> apply_2#(x1,y)
p2: apply_2#(f,x) -> app#(f,x)

and R consists of:

r1: app(apply_0(),y) -> apply_1(y)
r2: app(apply_1(x1),y) -> apply_2(x1,y)
r3: apply_2(f,x) -> app(f,x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 3, x2 - 1}
            apply_1_A(x1) = max{2, x1}
            apply_2#_A(x1,x2) = max{0, x1 - 3, x2 - 1}
            app#!_A(x1,x2) = max{2, x1 - 3, x2 - 1}
            apply_1!_A(x1) = max{1, x1 - 3}
            apply_2#!_A(x1,x2) = max{2, x1 - 3, x2 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 8, x2 - 1}
            apply_1_A(x1) = x1 + 7
            apply_2#_A(x1,x2) = max{0, x1 - 8, x2 - 1}
            app#!_A(x1,x2) = max{3, x1 - 3, x2 - 1}
            apply_1!_A(x1) = max{2, x1 - 4}
            apply_2#!_A(x1,x2) = max{3, x1 - 3, x2 - 1}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(apply_1) = [1]
      pi(apply_2#) = [1]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: apply_2#(f,x) -> app#(f,x)

and R consists of:

r1: app(apply_0(),y) -> apply_1(y)
r2: app(apply_1(x1),y) -> apply_2(x1,y)
r3: apply_2(f,x) -> app(f,x)

The estimated dependency graph contains the following SCCs:

  (no SCCs)