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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      shuffle#_A(x1) = x1
      add_A(x1,x2) = ((1,0,0),(1,0,0),(1,0,0)) x1 + ((1,0,0),(1,0,0),(0,1,0)) x2 + (4,4,1)
      reverse_A(x1) = ((1,0,0),(1,1,0),(0,0,1)) x1 + (1,3,1)
      app_A(x1,x2) = ((1,0,0),(1,0,0),(0,0,0)) x1 + ((1,0,0),(1,1,0),(0,0,0)) x2 + (0,1,3)
      nil_A() = (0,1,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      reverse#_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1
      add_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0),(0,1,0),(0,1,1)) x1 + ((0,0,0),(1,0,0),(0,1,0)) x2
      add_A(x1,x2) = ((1,0,0),(1,1,0),(0,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)

The next rules are strictly ordered:

  p1

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