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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quot#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
            s_A(x1) = x1 + (0,2)
            minus_A(x1,x2) = x1
            |0|_A() = (1,2)
            quot_A(x1,x2) = x1 + (0,1)
            app_A(x1,x2) = x2 + (3,4)
            nil_A() = (1,1)
            add_A(x1,x2) = (2,3)
            reverse_A(x1) = ((1,0),(0,0)) x1 + (4,2)
            shuffle_A(x1) = (5,4)
            concat_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (5,10)
            leaf_A() = (2,2)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (7,6)
            less_leaves_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,2)
            false_A() = (1,1)
            true_A() = (1,5)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,3)
    
      precedence:
      
        less_leaves > quot# = app = nil = add = reverse = shuffle = concat = cons > minus > quot > s = |0| = leaf = false = true = rand
      
      partial status:
    
        pi(quot#) = [1, 2]
        pi(s) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(quot) = []
        pi(app) = [2]
        pi(nil) = []
        pi(add) = []
        pi(reverse) = []
        pi(shuffle) = []
        pi(concat) = [1, 2]
        pi(leaf) = []
        pi(cons) = [1, 2]
        pi(less_leaves) = []
        pi(false) = []
        pi(true) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            quot#_A(x1,x2) = x1 + (3,2)
            s_A(x1) = x1
            minus_A(x1,x2) = x1 + (2,1)
            |0|_A() = (0,0)
            quot_A(x1,x2) = (1,1)
            app_A(x1,x2) = ((1,0),(1,1)) x2 + (3,0)
            nil_A() = (1,0)
            add_A(x1,x2) = (2,1)
            reverse_A(x1) = (3,0)
            shuffle_A(x1) = (4,2)
            concat_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,2)
            leaf_A() = (1,0)
            cons_A(x1,x2) = x2 + (0,1)
            less_leaves_A(x1,x2) = (0,0)
            false_A() = (0,0)
            true_A() = (0,0)
            rand_A(x1) = (1,1)
    
      precedence:
      
        minus = quot > concat = leaf = cons = less_leaves = false = true > app > rand > quot# = s = |0| = nil = add = reverse = shuffle
      
      partial status:
    
        pi(quot#) = [1]
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(quot) = []
        pi(app) = []
        pi(nil) = []
        pi(add) = []
        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: 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:

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

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            shuffle#_A(x1) = ((1,0),(1,1)) x1 + (1,3)
            add_A(x1,x2) = ((1,0),(0,0)) x2 + (8,3)
            reverse_A(x1) = ((1,0),(1,0)) x1 + (0,1)
            minus_A(x1,x2) = x1
            |0|_A() = (0,2)
            s_A(x1) = x1 + (0,1)
            quot_A(x1,x2) = x1 + (1,2)
            app_A(x1,x2) = x1 + x2 + (0,4)
            nil_A() = (0,0)
            shuffle_A(x1) = x1 + (1,1)
            concat_A(x1,x2) = x1 + x2 + (1,1)
            leaf_A() = (2,4)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (4,4)
            less_leaves_A(x1,x2) = x1 + (2,2)
            false_A() = (1,5)
            true_A() = (1,5)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
    
      precedence:
      
        minus > |0| > shuffle# > less_leaves > concat > cons > reverse = nil = shuffle = leaf = false = rand > add = app > quot > true > s
      
      partial status:
    
        pi(shuffle#) = [1]
        pi(add) = []
        pi(reverse) = []
        pi(minus) = [1]
        pi(|0|) = []
        pi(s) = [1]
        pi(quot) = [1]
        pi(app) = [1]
        pi(nil) = []
        pi(shuffle) = [1]
        pi(concat) = [1, 2]
        pi(leaf) = []
        pi(cons) = [1, 2]
        pi(less_leaves) = []
        pi(false) = []
        pi(true) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            shuffle#_A(x1) = (0,0)
            add_A(x1,x2) = (2,2)
            reverse_A(x1) = (4,8)
            minus_A(x1,x2) = x1 + (0,4)
            |0|_A() = (1,1)
            s_A(x1) = ((1,0),(1,0)) x1 + (2,2)
            quot_A(x1,x2) = x1 + (0,1)
            app_A(x1,x2) = ((0,0),(1,0)) x1 + (3,3)
            nil_A() = (1,1)
            shuffle_A(x1) = (5,9)
            concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
            leaf_A() = (0,0)
            cons_A(x1,x2) = ((1,0),(1,0)) x1 + (2,2)
            less_leaves_A(x1,x2) = (0,0)
            false_A() = (0,0)
            true_A() = (0,0)
            rand_A(x1) = (0,0)
    
      precedence:
      
        shuffle = cons > concat > rand > reverse > shuffle# = add = minus = |0| = s = quot = app = nil = leaf = less_leaves = false = true
      
      partial status:
    
        pi(shuffle#) = []
        pi(add) = []
        pi(reverse) = []
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = [1]
        pi(app) = []
        pi(nil) = []
        pi(shuffle) = []
        pi(concat) = [1, 2]
        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: 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            reverse#_A(x1) = ((1,0),(1,1)) x1 + (1,4)
            add_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (7,3)
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (1,2)
            |0|_A() = (0,0)
            s_A(x1) = ((1,0),(0,0)) x1
            quot_A(x1,x2) = ((0,0),(1,0)) x2 + (2,1)
            app_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (0,4)
            nil_A() = (0,2)
            reverse_A(x1) = ((1,0),(1,1)) x1
            shuffle_A(x1) = ((1,0),(1,0)) x1 + (8,1)
            concat_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,3)
            leaf_A() = (2,4)
            cons_A(x1,x2) = x1 + x2 + (4,1)
            less_leaves_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,1)
            false_A() = (1,4)
            true_A() = (0,3)
            rand_A(x1) = x1 + (1,1)
    
      precedence:
      
        minus = nil = reverse = shuffle = rand > reverse# = app > add = concat > quot = leaf = less_leaves = false = true > |0| > s = cons
      
      partial status:
    
        pi(reverse#) = [1]
        pi(add) = [2]
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = []
        pi(app) = [2]
        pi(nil) = []
        pi(reverse) = []
        pi(shuffle) = []
        pi(concat) = [1]
        pi(leaf) = []
        pi(cons) = [1, 2]
        pi(less_leaves) = []
        pi(false) = []
        pi(true) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            reverse#_A(x1) = ((1,0),(1,1)) x1 + (1,2)
            add_A(x1,x2) = x2 + (2,1)
            minus_A(x1,x2) = (1,1)
            |0|_A() = (0,0)
            s_A(x1) = (2,2)
            quot_A(x1,x2) = (3,3)
            app_A(x1,x2) = x2 + (1,2)
            nil_A() = (1,0)
            reverse_A(x1) = (5,4)
            shuffle_A(x1) = (6,5)
            concat_A(x1,x2) = ((1,0),(1,1)) x1 + (1,0)
            leaf_A() = (0,0)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (2,1)
            less_leaves_A(x1,x2) = (3,2)
            false_A() = (0,0)
            true_A() = (0,0)
            rand_A(x1) = (0,0)
    
      precedence:
      
        minus > app = reverse = shuffle > concat > reverse# > nil > rand > add = |0| = s = quot = leaf = cons = less_leaves = false = true
      
      partial status:
    
        pi(reverse#) = [1]
        pi(add) = []
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = []
        pi(app) = [2]
        pi(nil) = []
        pi(reverse) = []
        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: 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:

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

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          less_leaves#_A(x1,x2) = max{8, x1 + 2, x2 + 1}
          cons_A(x1,x2) = max{x1 + 6, x2}
          concat_A(x1,x2) = max{x1 + 5, x2}
          minus_A(x1,x2) = x1 + 5
          |0|_A = 0
          s_A(x1) = x1
          quot_A(x1,x2) = x2 + 3
          app_A(x1,x2) = max{20, x1 - 9, x2}
          nil_A = 0
          add_A(x1,x2) = max{9, x2 - 2}
          reverse_A(x1) = x1 + 12
          shuffle_A(x1) = 10
          leaf_A = 4
          less_leaves_A(x1,x2) = max{x1 + 6, x2}
          false_A = 1
          true_A = 3
          rand_A(x1) = x1 + 2
    
      precedence:
      
        cons = concat > |0| > minus = s = quot = app = shuffle = leaf > true > add = reverse > less_leaves# > nil = less_leaves = false > rand
      
      partial status:
    
        pi(less_leaves#) = [1, 2]
        pi(cons) = [1, 2]
        pi(concat) = [1, 2]
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = [2]
        pi(app) = [2]
        pi(nil) = []
        pi(add) = []
        pi(reverse) = []
        pi(shuffle) = []
        pi(leaf) = []
        pi(less_leaves) = [2]
        pi(false) = []
        pi(true) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          less_leaves#_A(x1,x2) = max{x1 + 1, x2 + 1}
          cons_A(x1,x2) = max{x1 + 2, x2}
          concat_A(x1,x2) = max{x1, x2}
          minus_A(x1,x2) = 18
          |0|_A = 14
          s_A(x1) = 8
          quot_A(x1,x2) = x2 + 10
          app_A(x1,x2) = max{12, x2}
          nil_A = 16
          add_A(x1,x2) = 1
          reverse_A(x1) = 11
          shuffle_A(x1) = 4
          leaf_A = 0
          less_leaves_A(x1,x2) = max{0, x2 - 3}
          false_A = 1
          true_A = 1
          rand_A(x1) = 7
    
      precedence:
      
        less_leaves# = cons = concat = true > minus = |0| = s = quot = app = nil = add = reverse = shuffle = leaf = less_leaves = false = rand
      
      partial status:
    
        pi(less_leaves#) = [2]
        pi(cons) = [1, 2]
        pi(concat) = [1]
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = [2]
        pi(app) = [2]
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          concat#_A(x1,x2) = max{x1 + 1, x2 + 1}
          cons_A(x1,x2) = max{x1 + 7, x2}
          minus_A(x1,x2) = max{x1 + 3, x2 - 1}
          |0|_A = 0
          s_A(x1) = x1
          quot_A(x1,x2) = x2 + 4
          app_A(x1,x2) = max{5, x1, x2 + 2}
          nil_A = 8
          add_A(x1,x2) = 1
          reverse_A(x1) = 11
          shuffle_A(x1) = 11
          concat_A(x1,x2) = max{7, x1 + 4, x2}
          leaf_A = 6
          less_leaves_A(x1,x2) = max{x1 - 4, x2 + 8}
          false_A = 1
          true_A = 1
          rand_A(x1) = x1 + 2
    
      precedence:
      
        concat# > false = true > cons = minus = |0| = quot = app = nil = add = reverse = shuffle = concat = leaf = less_leaves > s = rand
      
      partial status:
    
        pi(concat#) = [1]
        pi(cons) = [1, 2]
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = [2]
        pi(app) = []
        pi(nil) = []
        pi(add) = []
        pi(reverse) = []
        pi(shuffle) = []
        pi(concat) = [1, 2]
        pi(leaf) = []
        pi(less_leaves) = []
        pi(false) = []
        pi(true) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          concat#_A(x1,x2) = max{6, x1 + 1}
          cons_A(x1,x2) = max{5, x1 + 3, x2}
          minus_A(x1,x2) = 10
          |0|_A = 10
          s_A(x1) = 4
          quot_A(x1,x2) = max{9, x2 + 1}
          app_A(x1,x2) = 33
          nil_A = 44
          add_A(x1,x2) = 11
          reverse_A(x1) = 33
          shuffle_A(x1) = 16
          concat_A(x1,x2) = max{x1 + 7, x2 + 7}
          leaf_A = 0
          less_leaves_A(x1,x2) = 0
          false_A = 1
          true_A = 2
          rand_A(x1) = 0
    
      precedence:
      
        concat# > false > |0| = quot = nil = reverse = less_leaves > cons = minus = s = app = add = shuffle = concat = leaf = true > rand
      
      partial status:
    
        pi(concat#) = [1]
        pi(cons) = [1, 2]
        pi(minus) = []
        pi(|0|) = []
        pi(s) = []
        pi(quot) = [2]
        pi(app) = []
        pi(nil) = []
        pi(add) = []
        pi(reverse) = []
        pi(shuffle) = []
        pi(concat) = [1, 2]
        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.