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:

  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:
          
            shuffle#_A(x1) = max{0, x1 - 3}
            add_A(x1,x2) = max{5, x1 - 3, x2}
            reverse_A(x1) = max{2, x1}
            app_A(x1,x2) = max{3, x1, x2}
            nil_A = 3
            shuffle#!_A(x1) = x1 + 2
            add!_A(x1,x2) = 0
            reverse!_A(x1) = max{8, x1 - 6}
            app!_A(x1,x2) = max{1, x1 - 9, x2 - 10}
            nil!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              shuffle#_A(x1) = x1
              add_A(x1,x2) = x1 + x2 + 2
              reverse_A(x1) = x1 + 1
              app_A(x1,x2) = x1 + x2
              nil_A() = 0
              shuffle#!_A(x1) = x1
              add!_A(x1,x2) = x2 + 4
              reverse!_A(x1) = 3
              app!_A(x1,x2) = 2
              nil!_A() = 1
        
  
    partial status:
  
      pi(shuffle#) = []
      pi(add) = []
      pi(reverse) = [1]
      pi(app) = [2]
      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: 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:

  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:
          
            reverse#_A(x1) = 0
            add_A(x1,x2) = max{x1 + 18, x2 - 3}
            reverse#!_A(x1) = 0
            add!_A(x1,x2) = max{x1 + 11, x2 - 51}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              reverse#_A(x1) = x1
              add_A(x1,x2) = x1 + x2 + 2
              reverse#!_A(x1) = x1 + 4
              add!_A(x1,x2) = x1 + x2 + 3
        
  
    partial status:
  
      pi(reverse#) = []
      pi(add) = [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: 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:

  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) = 0
            add_A(x1,x2) = max{18, x1 - 8, x2 - 15}
            app#!_A(x1,x2) = max{9, x2 - 1}
            add!_A(x1,x2) = max{8, x2 + 5}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              add_A(x1,x2) = x1 + x2 + 2
              app#!_A(x1,x2) = x1 + 1
              add!_A(x1,x2) = 2
        
  
    partial status:
  
      pi(app#) = []
      pi(add) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.