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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      sum#_A(x1) = ((1,0,0),(0,1,0),(1,1,0)) x1
      plus_A(x1,x2) = ((1,0,0),(1,0,0),(1,0,0)) x1 + ((1,0,0),(1,1,0),(1,0,0)) x2 + (3,1,1)
      cons_A(x1,x2) = x1 + x2 + (4,11,1)
      |0|_A() = (0,0,1)
      s_A(x1) = x1 + (2,1,1)
      app_A(x1,x2) = ((1,0,0),(0,1,0),(1,0,0)) x1 + ((1,0,0),(1,0,0),(0,1,0)) x2 + (14,1,0)
      sum_A(x1) = ((1,0,0),(0,0,0),(1,0,0)) x1 + (0,13,0)
      nil_A() = (0,1,4)
      pred_A(x1) = ((1,0,0),(1,0,0),(0,1,0)) x1 + (0,7,0)

The next rules are strictly ordered:

  p1, p2, p4

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      sum#_A(x1) = ((1,0,0),(0,0,0),(0,1,0)) x1
      app_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,1)) x1 + ((1,0,0),(1,0,0),(0,0,0)) x2 + (1,1,1)
      cons_A(x1,x2) = ((0,0,0),(1,0,0),(1,0,0)) x1 + ((1,0,0),(0,1,0),(1,0,1)) x2 + (5,1,22)
      sum_A(x1) = ((0,0,0),(1,0,0),(0,0,0)) x1 + (9,2,22)
      nil_A() = (1,6,0)
      plus_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,2,2)
      |0|_A() = (0,1,1)
      s_A(x1) = x1 + (0,1,1)
      pred_A(x1) = ((0,0,0),(1,0,0),(1,1,0)) x1 + (8,0,0)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,0)) x1 + ((1,0,0),(0,0,0),(0,1,0)) x2
      cons_A(x1,x2) = x2 + (2,3,6)
      app_A(x1,x2) = x1 + x2 + (1,0,7)
      nil_A() = (0,1,1)
      sum_A(x1) = ((0,0,0),(1,0,0),(0,1,0)) x1 + (3,0,0)
      plus_A(x1,x2) = ((1,0,0),(0,1,0),(1,0,1)) x2 + (5,1,0)
      |0|_A() = (1,1,0)
      s_A(x1) = x1
      pred_A(x1) = ((1,0,0),(1,0,0),(0,1,0)) x1 + (0,3,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      plus#_A(x1,x2) = x1
      s_A(x1) = ((1,0,0),(0,1,0),(1,0,0)) x1 + (7,0,0)
      app_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,0)) x1 + ((1,0,0),(1,0,0),(1,1,0)) x2 + (0,1,0)
      nil_A() = (1,1,0)
      cons_A(x1,x2) = ((0,0,0),(1,0,0),(1,0,0)) x1 + ((1,0,0),(0,1,0),(0,1,0)) x2 + (3,0,1)
      sum_A(x1) = ((0,0,0),(1,0,0),(0,1,0)) x1 + (5,0,2)
      plus_A(x1,x2) = ((1,0,0),(0,1,0),(1,0,0)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (6,1,0)
      |0|_A() = (1,1,0)
      pred_A(x1) = ((1,0,0),(1,1,0),(0,0,1)) x1

The next rules are strictly ordered:

  p1

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