YES

We show the termination of the TRS R:

  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  quot(|0|(),s(y)) -> |0|()
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  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)))
  concat(leaf(),y) -> y
  concat(cons(u,v),y) -> cons(u,concat(v,y))
  less_leaves(x,leaf()) -> false()
  less_leaves(leaf(),cons(w,z)) -> true()
  less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

-- SCC decomposition.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)
p2: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p3: quot#(s(x),s(y)) -> minus#(x,y)
p4: app#(add(n,x),y) -> app#(x,y)
p5: reverse#(add(n,x)) -> app#(reverse(x),add(n,nil()))
p6: reverse#(add(n,x)) -> reverse#(x)
p7: shuffle#(add(n,x)) -> shuffle#(reverse(x))
p8: shuffle#(add(n,x)) -> reverse#(x)
p9: concat#(cons(u,v),y) -> concat#(v,y)
p10: less_leaves#(cons(u,v),cons(w,z)) -> less_leaves#(concat(u,v),concat(w,z))
p11: less_leaves#(cons(u,v),cons(w,z)) -> concat#(u,v)
p12: less_leaves#(cons(u,v),cons(w,z)) -> concat#(w,z)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: app(nil(),y) -> y
r6: app(add(n,x),y) -> add(n,app(x,y))
r7: reverse(nil()) -> nil()
r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r9: shuffle(nil()) -> nil()
r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))
r11: concat(leaf(),y) -> y
r12: concat(cons(u,v),y) -> cons(u,concat(v,y))
r13: less_leaves(x,leaf()) -> false()
r14: less_leaves(leaf(),cons(w,z)) -> true()
r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}
  {p7}
  {p6}
  {p4}
  {p10}
  {p9}


-- Reduction pair.

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

p1: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: app(nil(),y) -> y
r6: app(add(n,x),y) -> add(n,app(x,y))
r7: reverse(nil()) -> nil()
r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r9: shuffle(nil()) -> nil()
r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))
r11: concat(leaf(),y) -> y
r12: concat(cons(u,v),y) -> cons(u,concat(v,y))
r13: less_leaves(x,leaf()) -> false()
r14: less_leaves(leaf(),cons(w,z)) -> true()
r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        quot#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 2
        minus_A(x1,x2) = x1 + 1
        |0|_A() = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        quot#_A(x1,x2) = x2
        s_A(x1) = x1 + 1
        minus_A(x1,x2) = x1 + 2
        |0|_A() = 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: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: app(nil(),y) -> y
r6: app(add(n,x),y) -> add(n,app(x,y))
r7: reverse(nil()) -> nil()
r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r9: shuffle(nil()) -> nil()
r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))
r11: concat(leaf(),y) -> y
r12: concat(cons(u,v),y) -> cons(u,concat(v,y))
r13: less_leaves(x,leaf()) -> false()
r14: less_leaves(leaf(),cons(w,z)) -> true()
r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        minus#_A(x1,x2) = 0
        s_A(x1) = x1 + 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: shuffle#(add(n,x)) -> shuffle#(reverse(x))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: app(nil(),y) -> y
r6: app(add(n,x),y) -> add(n,app(x,y))
r7: reverse(nil()) -> nil()
r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r9: shuffle(nil()) -> nil()
r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))
r11: concat(leaf(),y) -> y
r12: concat(cons(u,v),y) -> cons(u,concat(v,y))
r13: less_leaves(x,leaf()) -> false()
r14: less_leaves(leaf(),cons(w,z)) -> true()
r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

The set of usable rules consists of

  r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      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
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        shuffle#_A(x1) = x1
        add_A(x1,x2) = 1
        reverse_A(x1) = x1 + 3
        app_A(x1,x2) = x2 + 2
        nil_A() = 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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: app(nil(),y) -> y
r6: app(add(n,x),y) -> add(n,app(x,y))
r7: reverse(nil()) -> nil()
r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r9: shuffle(nil()) -> nil()
r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))
r11: concat(leaf(),y) -> y
r12: concat(cons(u,v),y) -> cons(u,concat(v,y))
r13: less_leaves(x,leaf()) -> false()
r14: less_leaves(leaf(),cons(w,z)) -> true()
r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        reverse#_A(x1) = x1
        add_A(x1,x2) = x1 + x2 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        reverse#_A(x1) = x1
        add_A(x1,x2) = x1 + x2 + 1
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: app(nil(),y) -> y
r6: app(add(n,x),y) -> add(n,app(x,y))
r7: reverse(nil()) -> nil()
r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r9: shuffle(nil()) -> nil()
r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))
r11: concat(leaf(),y) -> y
r12: concat(cons(u,v),y) -> cons(u,concat(v,y))
r13: less_leaves(x,leaf()) -> false()
r14: less_leaves(leaf(),cons(w,z)) -> true()
r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        add_A(x1,x2) = x1 + x2 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x2
        add_A(x1,x2) = x1 + x2 + 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: less_leaves#(cons(u,v),cons(w,z)) -> less_leaves#(concat(u,v),concat(w,z))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: app(nil(),y) -> y
r6: app(add(n,x),y) -> add(n,app(x,y))
r7: reverse(nil()) -> nil()
r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r9: shuffle(nil()) -> nil()
r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))
r11: concat(leaf(),y) -> y
r12: concat(cons(u,v),y) -> cons(u,concat(v,y))
r13: less_leaves(x,leaf()) -> false()
r14: less_leaves(leaf(),cons(w,z)) -> true()
r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

The set of usable rules consists of

  r11, r12

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        less_leaves#_A(x1,x2) = x1 + x2
        cons_A(x1,x2) = x1 + x2 + 2
        concat_A(x1,x2) = x1 + x2 + 1
        leaf_A() = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        less_leaves#_A(x1,x2) = x1 + x2
        cons_A(x1,x2) = 1
        concat_A(x1,x2) = x2 + 2
        leaf_A() = 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: concat#(cons(u,v),y) -> concat#(v,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: app(nil(),y) -> y
r6: app(add(n,x),y) -> add(n,app(x,y))
r7: reverse(nil()) -> nil()
r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil()))
r9: shuffle(nil()) -> nil()
r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x)))
r11: concat(leaf(),y) -> y
r12: concat(cons(u,v),y) -> cons(u,concat(v,y))
r13: less_leaves(x,leaf()) -> false()
r14: less_leaves(leaf(),cons(w,z)) -> true()
r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        concat#_A(x1,x2) = x1 + x2
        cons_A(x1,x2) = x1 + x2 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        concat#_A(x1,x2) = x2
        cons_A(x1,x2) = x1 + x2 + 1
    

The next rules are strictly ordered:

  p1

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