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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        shuffle#_A(x1) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,1,0)) x1
        add_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,1)) x2 + (3,1,1,1)
        reverse_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,1,1)) x1 + (2,3,3,1)
        app_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,1,1)) x2 + (0,1,1,1)
        nil_A() = (0,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        shuffle# > app > nil > reverse > add
      
      argument filter:
    
        pi(shuffle#) = []
        pi(add) = 2
        pi(reverse) = [1]
        pi(app) = 1
        pi(nil) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        reverse#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x1
        add_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        add > reverse#
      
      argument filter:
    
        pi(reverse#) = []
        pi(add) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        app#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x2
        add_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,1,0,0)) x2 + (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        add > app#
      
      argument filter:
    
        pi(app#) = []
        pi(add) = []
    

The next rules are strictly ordered:

  p1

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