YES
We show the termination of the TRS R:

  app(nil(),y) -> y
  app(add(n,x),y) -> add(n,app(x,y))
  reverse(nil()) -> nil()
  reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
  shuffle(nil()) -> nil()
  shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))


-- SCC decomposition.

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

p1: app#(add(n,x),y) -> app#(x,y)
p2: reverse#(add(n,x)) -> app#(reverse(x),add(n,nil()))
p3: reverse#(add(n,x)) -> reverse#(x)
p4: shuffle#(add(n,x)) -> shuffle#(reverse(x))
p5: shuffle#(add(n,x)) -> reverse#(x)

and R consists of:

r1: app(nil(),y) -> y
r2: app(add(n,x),y) -> add(n,app(x,y))
r3: reverse(nil()) -> nil()
r4: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r5: shuffle(nil()) -> nil()
r6: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p3}
  {p1}


-- Reduction pair.

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

p1: shuffle#(add(n,x)) -> shuffle#(reverse(x))

and R consists of:

r1: app(nil(),y) -> y
r2: app(add(n,x),y) -> add(n,app(x,y))
r3: reverse(nil()) -> nil()
r4: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r5: shuffle(nil()) -> nil()
r6: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the monotone 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:
          shuffle#_A(x1) = x1
          add_A(x1,x2) = x1 + x2 + 3
          reverse_A(x1) = x1 + 1
          app_A(x1,x2) = x1 + x2
          nil_A() = 0
          shuffle_A(x1) = x1 + 1
  
    base order 2:
  
      precedence:
      
        shuffle > reverse > app > add > shuffle# = nil
      
  
    partial status:
  
      pi(shuffle#) = [1]
      pi(add) = [1, 2]
      pi(reverse) = [1]
      pi(app) = [1, 2]
      pi(nil) = []
      pi(shuffle) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: reverse#(add(n,x)) -> reverse#(x)

and R consists of:

r1: app(nil(),y) -> y
r2: app(add(n,x),y) -> add(n,app(x,y))
r3: reverse(nil()) -> nil()
r4: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r5: shuffle(nil()) -> nil()
r6: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))

The set of usable rules consists of

  r5

Take the monotone 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:
          reverse#_A(x1) = x1
          add_A(x1,x2) = x1 + x2 + 3
          shuffle_A(x1) = x1 + 1
          nil_A() = 2
  
    base order 2:
  
      precedence:
      
        add = shuffle > reverse# > nil
      
  
    partial status:
  
      pi(reverse#) = [1]
      pi(add) = [1, 2]
      pi(shuffle) = [1]
      pi(nil) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: app#(add(n,x),y) -> app#(x,y)

and R consists of:

r1: app(nil(),y) -> y
r2: app(add(n,x),y) -> add(n,app(x,y))
r3: reverse(nil()) -> nil()
r4: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r5: shuffle(nil()) -> nil()
r6: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))

The set of usable rules consists of

  r5

Take the monotone 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 + x2 + 4
          add_A(x1,x2) = x1 + x2 + 3
          shuffle_A(x1) = x1 + 1
          nil_A() = 2
  
    base order 2:
  
      precedence:
      
        app# > add > shuffle > nil
      
  
    partial status:
  
      pi(app#) = [1, 2]
      pi(add) = [1, 2]
      pi(shuffle) = [1]
      pi(nil) = []

The following rules are removed:

  p1

Then no dependency pair remains.