YES

We show the termination of the TRS 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())

-- SCC decomposition.

Consider the 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())

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1}
  {p2}
  {p7}


-- Reduction pair.

Consider the 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())

The set of usable rules consists of

  r1, r2, r3, r4, r5, r7, r8

Take the reduction pair:

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

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: 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())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = x1 + 1
          cons_A(x1,x2) = max{x1, x2 + 1}
    
      precedence:
      
        app# = cons
      
      partial status:
    
        pi(app#) = [1]
        pi(cons) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 1}
          cons_A(x1,x2) = max{x1, x2}
    
      precedence:
      
        app# = cons
      
      partial status:
    
        pi(app#) = []
        pi(cons) = [1, 2]
    

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: 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())

The set of usable rules consists of

  r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = x1 + 9
          cons_A(x1,x2) = x2 + 2
          plus_A(x1,x2) = max{x1 + 8, x2 + 8}
          |0|_A = 0
          s_A(x1) = max{7, x1}
    
      precedence:
      
        sum# = cons = plus > |0| = s
      
      partial status:
    
        pi(sum#) = []
        pi(cons) = [2]
        pi(plus) = [1, 2]
        pi(|0|) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = 19
          cons_A(x1,x2) = max{11, x2 + 7}
          plus_A(x1,x2) = max{12, x1 - 2, x2 + 1}
          |0|_A = 0
          s_A(x1) = max{13, x1 + 1}
    
      precedence:
      
        sum# = cons = plus = |0| = s
      
      partial status:
    
        pi(sum#) = []
        pi(cons) = [2]
        pi(plus) = []
        pi(|0|) = []
        pi(s) = [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: 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())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{2, x1 + 1, x2}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        plus# = s
      
      partial status:
    
        pi(plus#) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{x1 - 1, x2 + 1}
          s_A(x1) = x1
    
      precedence:
      
        plus# = s
      
      partial status:
    
        pi(plus#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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