YES

We show the termination of the relative TRS R/S:

  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))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

Consider the non-minimal 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))
r16: rand(x) -> x
r17: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

Consider the non-minimal 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))
r16: rand(x) -> x
r17: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          quot#_A(x1,x2) = x1
          s_A(x1) = x1
          minus_A(x1,x2) = x1
          |0|_A() = (0,0)
          quot_A(x1,x2) = x1 + (1,1)
          app_A(x1,x2) = x2 + (3,2)
          nil_A() = (0,5)
          add_A(x1,x2) = (3,1)
          reverse_A(x1) = ((1,0),(1,1)) x1 + (4,0)
          shuffle_A(x1) = x1 + (2,0)
          concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,8)
          leaf_A() = (1,1)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (3,1)
          less_leaves_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1)
          false_A() = (2,2)
          true_A() = (2,8)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,1)
  
    precedence:
    
      leaf = less_leaves > concat = cons > reverse > app = nil = add = shuffle = false = true = rand > quot > s > quot# = minus = |0|
    
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(quot) = [1]
      pi(app) = [2]
      pi(nil) = []
      pi(add) = []
      pi(reverse) = [1]
      pi(shuffle) = []
      pi(concat) = [1]
      pi(leaf) = []
      pi(cons) = [1]
      pi(less_leaves) = []
      pi(false) = []
      pi(true) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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))
r16: rand(x) -> x
r17: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          minus#_A(x1,x2) = x1 + x2 + (1,2)
          s_A(x1) = x1 + (0,1)
          minus_A(x1,x2) = x1
          |0|_A() = (1,2)
          quot_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (1,2)
          app_A(x1,x2) = x2 + (1,1)
          nil_A() = (3,3)
          add_A(x1,x2) = (0,0)
          reverse_A(x1) = x1 + (2,2)
          shuffle_A(x1) = x1 + (1,1)
          concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,6)
          leaf_A() = (2,2)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
          less_leaves_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
          false_A() = (1,1)
          true_A() = (0,0)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,2)
  
    precedence:
    
      minus = reverse = shuffle = concat = rand > nil = leaf = cons > minus# = app > add > s = |0| = quot = less_leaves = false = true
    
    partial status:
  
      pi(minus#) = [1]
      pi(s) = []
      pi(minus) = []
      pi(|0|) = []
      pi(quot) = []
      pi(app) = [2]
      pi(nil) = []
      pi(add) = []
      pi(reverse) = []
      pi(shuffle) = []
      pi(concat) = []
      pi(leaf) = []
      pi(cons) = [1, 2]
      pi(less_leaves) = [1, 2]
      pi(false) = []
      pi(true) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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))
r16: rand(x) -> x
r17: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          shuffle#_A(x1) = x1
          add_A(x1,x2) = x2 + (0,1)
          reverse_A(x1) = x1
          minus_A(x1,x2) = x1
          |0|_A() = (1,2)
          s_A(x1) = x1 + (0,1)
          quot_A(x1,x2) = x1
          app_A(x1,x2) = x1 + ((1,0),(1,1)) x2
          nil_A() = (0,0)
          shuffle_A(x1) = x1 + (1,2)
          concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,4)
          leaf_A() = (2,2)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
          less_leaves_A(x1,x2) = x1 + x2 + (1,1)
          false_A() = (1,1)
          true_A() = (1,3)
          rand_A(x1) = ((1,0),(1,0)) x1 + (1,2)
  
    precedence:
    
      reverse = s = leaf > less_leaves > concat > nil > quot > minus = app = shuffle > cons > shuffle# = add = |0| = false = true = rand
    
    partial status:
  
      pi(shuffle#) = []
      pi(add) = []
      pi(reverse) = []
      pi(minus) = [1]
      pi(|0|) = []
      pi(s) = []
      pi(quot) = [1]
      pi(app) = [2]
      pi(nil) = []
      pi(shuffle) = []
      pi(concat) = [2]
      pi(leaf) = []
      pi(cons) = [2]
      pi(less_leaves) = [1]
      pi(false) = []
      pi(true) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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))
r16: rand(x) -> x
r17: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          reverse#_A(x1) = ((1,0),(1,0)) x1 + (1,2)
          add_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (6,1)
          minus_A(x1,x2) = x1
          |0|_A() = (0,1)
          s_A(x1) = x1 + (0,2)
          quot_A(x1,x2) = x1 + (0,1)
          app_A(x1,x2) = x1 + x2 + (0,2)
          nil_A() = (0,3)
          reverse_A(x1) = ((1,0),(1,1)) x1
          shuffle_A(x1) = ((1,0),(1,0)) x1 + (7,2)
          concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,5)
          leaf_A() = (2,2)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (4,4)
          less_leaves_A(x1,x2) = x1 + x2 + (1,1)
          false_A() = (1,1)
          true_A() = (1,3)
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,3)
  
    precedence:
    
      minus = nil = reverse = leaf = true = rand > reverse# = add = app = shuffle > concat = cons = less_leaves > |0| = s = quot = false
    
    partial status:
  
      pi(reverse#) = []
      pi(add) = []
      pi(minus) = []
      pi(|0|) = []
      pi(s) = []
      pi(quot) = []
      pi(app) = [2]
      pi(nil) = []
      pi(reverse) = []
      pi(shuffle) = []
      pi(concat) = []
      pi(leaf) = []
      pi(cons) = []
      pi(less_leaves) = []
      pi(false) = []
      pi(true) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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))
r16: rand(x) -> x
r17: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((1,0),(1,0)) x1 + (5,2)
          add_A(x1,x2) = ((1,0),(0,0)) x2 + (4,1)
          minus_A(x1,x2) = x1 + (0,3)
          |0|_A() = (1,1)
          s_A(x1) = x1
          quot_A(x1,x2) = (2,2)
          app_A(x1,x2) = x1 + x2 + (0,2)
          nil_A() = (0,2)
          reverse_A(x1) = ((1,0),(1,0)) x1 + (0,3)
          shuffle_A(x1) = x1 + (5,1)
          concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,4)
          leaf_A() = (1,1)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
          less_leaves_A(x1,x2) = x1 + x2 + (2,1)
          false_A() = (2,3)
          true_A() = (0,0)
          rand_A(x1) = x1 + (1,1)
  
    precedence:
    
      s = quot = reverse > minus = shuffle = concat = leaf = false = true = rand > add = |0| = app = nil > cons = less_leaves > app#
    
    partial status:
  
      pi(app#) = []
      pi(add) = []
      pi(minus) = [1]
      pi(|0|) = []
      pi(s) = []
      pi(quot) = []
      pi(app) = [2]
      pi(nil) = []
      pi(reverse) = []
      pi(shuffle) = []
      pi(concat) = []
      pi(leaf) = []
      pi(cons) = []
      pi(less_leaves) = [1, 2]
      pi(false) = []
      pi(true) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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))
r16: rand(x) -> x
r17: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          less_leaves#_A(x1,x2) = x1 + (1,2)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (4,2)
          concat_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
          minus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,3)
          |0|_A() = (1,1)
          s_A(x1) = x1
          quot_A(x1,x2) = ((0,0),(1,0)) x2 + (2,2)
          app_A(x1,x2) = x2 + (0,1)
          nil_A() = (0,1)
          add_A(x1,x2) = (0,0)
          reverse_A(x1) = (0,2)
          shuffle_A(x1) = x1 + (0,3)
          leaf_A() = (6,2)
          less_leaves_A(x1,x2) = x2 + (2,2)
          false_A() = (1,1)
          true_A() = (5,5)
          rand_A(x1) = ((1,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      concat = minus > less_leaves# = cons = |0| = s = quot > app = nil = add = reverse = shuffle = leaf = less_leaves = rand > false = true
    
    partial status:
  
      pi(less_leaves#) = [1]
      pi(cons) = []
      pi(concat) = []
      pi(minus) = []
      pi(|0|) = []
      pi(s) = []
      pi(quot) = []
      pi(app) = []
      pi(nil) = []
      pi(add) = []
      pi(reverse) = []
      pi(shuffle) = []
      pi(leaf) = []
      pi(less_leaves) = []
      pi(false) = []
      pi(true) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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))
r16: rand(x) -> x
r17: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          concat#_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (3,1)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,2)
          minus_A(x1,x2) = ((1,0),(1,0)) x1 + (3,4)
          |0|_A() = (1,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          quot_A(x1,x2) = (2,3)
          app_A(x1,x2) = x2 + (2,2)
          nil_A() = (2,4)
          add_A(x1,x2) = ((0,0),(1,0)) x1 + (1,3)
          reverse_A(x1) = (4,1)
          shuffle_A(x1) = (5,2)
          concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
          leaf_A() = (2,2)
          less_leaves_A(x1,x2) = x1 + x2 + (1,1)
          false_A() = (1,1)
          true_A() = (0,0)
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
  
    precedence:
    
      concat# = cons = minus = app = nil = add = reverse = shuffle = concat = leaf = false = true = rand > less_leaves > |0| = s = quot
    
    partial status:
  
      pi(concat#) = []
      pi(cons) = []
      pi(minus) = []
      pi(|0|) = []
      pi(s) = []
      pi(quot) = []
      pi(app) = []
      pi(nil) = []
      pi(add) = []
      pi(reverse) = []
      pi(shuffle) = []
      pi(concat) = []
      pi(leaf) = []
      pi(less_leaves) = [2]
      pi(false) = []
      pi(true) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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