YES

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

  R:
  app(nil(),k) -> k
  app(l,nil()) -> l
  app(cons(x,l),k) -> cons(x,app(l,k))
  sum(cons(x,nil())) -> cons(x,nil())
  sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
  sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
  pred(cons(s(x),nil())) -> cons(x,nil())

  S:
  cons(x,cons(y,l)) -> cons(y,cons(x,l))

-- SCC decomposition.

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

p1: app#(cons(x,l),k) -> app#(l,k)
p2: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p3: sum#(cons(x,cons(y,l))) -> plus#(x,y)
p4: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p5: sum#(app(l,cons(x,cons(y,k)))) -> app#(l,sum(cons(x,cons(y,k))))
p6: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p7: plus#(s(x),y) -> plus#(x,y)
p8: sum#(plus(cons(|0|(),x),cons(y,l))) -> pred#(sum(cons(s(x),cons(y,l))))
p9: sum#(plus(cons(|0|(),x),cons(y,l))) -> sum#(cons(s(x),cons(y,l)))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: sum#(plus(cons(|0|(),x),cons(y,l))) -> sum#(cons(s(x),cons(y,l)))
p2: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p3: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p4: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          sum#_A(x1) = ((1,0),(1,0)) x1 + (0,1)
          plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (9,8)
          cons_A(x1,x2) = ((1,0),(0,0)) x2 + (4,19)
          |0|_A() = (7,21)
          s_A(x1) = x1 + (18,55)
          app_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (42,20)
          sum_A(x1) = ((1,0),(0,0)) x1 + (0,22)
          nil_A() = (1,1)
          pred_A(x1) = (6,20)
  
    precedence:
    
      sum# = plus = cons = |0| = s = app = sum = nil = pred
    
    partial status:
  
      pi(sum#) = []
      pi(plus) = [2]
      pi(cons) = []
      pi(|0|) = []
      pi(s) = []
      pi(app) = [1, 2]
      pi(sum) = []
      pi(nil) = []
      pi(pred) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p2: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p3: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p2: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p3: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          sum#_A(x1) = ((1,0),(0,0)) x1 + (1,8)
          app_A(x1,x2) = x1 + x2 + (2,4)
          cons_A(x1,x2) = ((0,0),(1,0)) x2 + (1,2)
          plus_A(x1,x2) = ((1,0),(0,0)) x2 + (3,9)
          sum_A(x1) = (1,6)
          nil_A() = (2,7)
          |0|_A() = (0,7)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          pred_A(x1) = (1,5)
  
    precedence:
    
      app = plus > |0| > sum# = cons = sum = nil = s = pred
    
    partial status:
  
      pi(sum#) = []
      pi(app) = []
      pi(cons) = []
      pi(plus) = []
      pi(sum) = []
      pi(nil) = []
      pi(|0|) = []
      pi(s) = []
      pi(pred) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p2: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p2: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          sum#_A(x1) = ((1,0),(0,0)) x1 + (0,10)
          cons_A(x1,x2) = ((1,0),(1,0)) x2 + (4,7)
          plus_A(x1,x2) = x2 + (9,12)
          app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (5,6)
          sum_A(x1) = ((1,0),(0,0)) x1 + (0,9)
          nil_A() = (1,10)
          |0|_A() = (7,8)
          s_A(x1) = (2,6)
          pred_A(x1) = x1 + (0,11)
  
    precedence:
    
      sum# = cons = app = sum > plus = nil = |0| = s > pred
    
    partial status:
  
      pi(sum#) = []
      pi(cons) = []
      pi(plus) = [2]
      pi(app) = [1]
      pi(sum) = []
      pi(nil) = []
      pi(|0|) = []
      pi(s) = []
      pi(pred) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          sum#_A(x1) = x1
          app_A(x1,x2) = x1 + x2 + (3,0)
          cons_A(x1,x2) = x2 + (2,4)
          sum_A(x1) = ((1,0),(0,0)) x1 + (0,8)
          nil_A() = (2,2)
          plus_A(x1,x2) = x2 + (5,9)
          |0|_A() = (8,9)
          s_A(x1) = (3,5)
          pred_A(x1) = (7,7)
  
    precedence:
    
      app > cons = sum = plus = pred > sum# = nil = |0| = s
    
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [2]
      pi(cons) = [2]
      pi(sum) = []
      pi(nil) = []
      pi(plus) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(pred) = []

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#(cons(x,l),k) -> app#(l,k)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((1,0),(1,1)) x1 + (1,5)
          cons_A(x1,x2) = ((1,0),(1,0)) x2 + (4,4)
          app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (5,6)
          nil_A() = (0,1)
          sum_A(x1) = (9,6)
          plus_A(x1,x2) = x2 + (10,9)
          |0|_A() = (2,3)
          s_A(x1) = (1,2)
          pred_A(x1) = (5,5)
  
    precedence:
    
      app = nil > app# = cons = sum = plus = |0| = s = pred
    
    partial status:
  
      pi(app#) = [1]
      pi(cons) = []
      pi(app) = [1]
      pi(nil) = []
      pi(sum) = []
      pi(plus) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(pred) = []

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: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))
r9: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r10: pred(cons(s(x),nil())) -> cons(x,nil())
r11: cons(x,cons(y,l)) -> cons(y,cons(x,l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = ((1,0),(1,1)) x1 + (0,1)
          s_A(x1) = x1 + (1,19)
          app_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (11,14)
          nil_A() = (1,13)
          cons_A(x1,x2) = ((1,0),(0,0)) x2 + (10,6)
          sum_A(x1) = ((0,0),(1,0)) x1 + (12,1)
          plus_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (9,22)
          |0|_A() = (0,20)
          pred_A(x1) = x1 + (0,8)
  
    precedence:
    
      s = nil = plus > plus# = pred > app = cons = sum = |0|
    
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(app) = [1, 2]
      pi(nil) = []
      pi(cons) = []
      pi(sum) = []
      pi(plus) = []
      pi(|0|) = []
      pi(pred) = []

The next rules are strictly ordered:

  p1

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