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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = max{48, x1 + 22}
          plus_A(x1,x2) = max{x1 + 19, x2 + 21}
          cons_A(x1,x2) = max{8, x2}
          |0|_A = 10
          s_A(x1) = max{64, x1}
          app_A(x1,x2) = max{63, x1 + 54, x2 + 45}
          sum_A(x1) = max{2, x1}
          nil_A = 0
          pred_A(x1) = max{16, x1 + 10}
    
      precedence:
      
        sum# = plus > |0| = s = app = sum > cons = nil = pred
      
      partial status:
    
        pi(sum#) = []
        pi(plus) = [1, 2]
        pi(cons) = []
        pi(|0|) = []
        pi(s) = [1]
        pi(app) = [1, 2]
        pi(sum) = []
        pi(nil) = []
        pi(pred) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = 172
          plus_A(x1,x2) = max{x1 - 158, x2 + 74}
          cons_A(x1,x2) = 232
          |0|_A = 307
          s_A(x1) = x1 + 157
          app_A(x1,x2) = max{x1 + 122, x2 + 111}
          sum_A(x1) = 66
          nil_A = 156
          pred_A(x1) = max{146, x1 + 2}
    
      precedence:
      
        sum# = plus = cons = |0| = s = app = sum = nil = pred
      
      partial status:
    
        pi(sum#) = []
        pi(plus) = [2]
        pi(cons) = []
        pi(|0|) = []
        pi(s) = [1]
        pi(app) = [2]
        pi(sum) = []
        pi(nil) = []
        pi(pred) = [1]
    

The next rules are strictly ordered:

  p1, p2

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)))))
p2: 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}


-- 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)))))
p2: 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1) = x1
            app_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (6,1)
            cons_A(x1,x2) = x1 + x2 + (5,2)
            sum_A(x1) = x1 + (0,4)
            plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (4,5)
            nil_A() = (7,1)
            |0|_A() = (4,3)
            s_A(x1) = x1 + (6,4)
            pred_A(x1) = x1 + (1,1)
    
      precedence:
      
        sum# = app = cons = sum = plus = nil = |0| = s = pred
      
      partial status:
    
        pi(sum#) = []
        pi(app) = []
        pi(cons) = []
        pi(sum) = []
        pi(plus) = []
        pi(nil) = []
        pi(|0|) = []
        pi(s) = []
        pi(pred) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1) = ((1,0),(0,0)) x1
            app_A(x1,x2) = ((1,0),(0,0)) x1 + (8,16)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (9,4)
            sum_A(x1) = ((1,0),(0,0)) x1 + (2,15)
            plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (6,16)
            nil_A() = (11,0)
            |0|_A() = (8,2)
            s_A(x1) = x1 + (10,1)
            pred_A(x1) = ((1,0),(0,0)) x1 + (1,3)
    
      precedence:
      
        sum = plus = pred > |0| > sum# = cons > app > nil = s
      
      partial status:
    
        pi(sum#) = []
        pi(app) = []
        pi(cons) = []
        pi(sum) = []
        pi(plus) = []
        pi(nil) = []
        pi(|0|) = []
        pi(s) = []
        pi(pred) = []
    

The next rules are strictly ordered:

  p2

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1) = ((1,0),(1,0)) x1 + (1,1)
            app_A(x1,x2) = x1 + x2 + (6,0)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (5,0)
            sum_A(x1) = (8,8)
            nil_A() = (1,9)
            plus_A(x1,x2) = ((1,0),(1,1)) x2 + (6,7)
            |0|_A() = (11,9)
            s_A(x1) = ((1,0),(1,0)) x1
            pred_A(x1) = (7,10)
    
      precedence:
      
        sum# = app = cons = sum = nil = plus = |0| = s = pred
      
      partial status:
    
        pi(sum#) = []
        pi(app) = []
        pi(cons) = []
        pi(sum) = []
        pi(nil) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
        pi(pred) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1) = (0,1)
            app_A(x1,x2) = x1 + x2
            cons_A(x1,x2) = (0,2)
            sum_A(x1) = (2,2)
            nil_A() = (2,2)
            plus_A(x1,x2) = x2 + (3,3)
            |0|_A() = (0,2)
            s_A(x1) = (1,3)
            pred_A(x1) = (1,3)
    
      precedence:
      
        sum# = app = cons = sum = nil = plus = pred > |0| = s
      
      partial status:
    
        pi(sum#) = []
        pi(app) = [2]
        pi(cons) = []
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1 + x2 + (10,16)
            cons_A(x1,x2) = x2 + (9,15)
            app_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,16)
            nil_A() = (7,1)
            sum_A(x1) = ((0,0),(1,0)) x1 + (17,1)
            plus_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (9,20)
            |0|_A() = (27,0)
            s_A(x1) = x1 + (26,24)
            pred_A(x1) = (17,2)
    
      precedence:
      
        app# = cons = app = nil = sum = plus = |0| = s = pred
      
      partial status:
    
        pi(app#) = []
        pi(cons) = []
        pi(app) = []
        pi(nil) = []
        pi(sum) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
        pi(pred) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1
            cons_A(x1,x2) = x2 + (0,6)
            app_A(x1,x2) = x2 + (4,1)
            nil_A() = (1,9)
            sum_A(x1) = (3,10)
            plus_A(x1,x2) = ((1,0),(0,0)) x2 + (3,13)
            |0|_A() = (1,5)
            s_A(x1) = ((0,0),(1,0)) x1 + (2,14)
            pred_A(x1) = (2,16)
    
      precedence:
      
        |0| = s > cons = app = sum = plus = pred > app# > nil
      
      partial status:
    
        pi(app#) = [1]
        pi(cons) = [2]
        pi(app) = [2]
        pi(nil) = []
        pi(sum) = []
        pi(plus) = []
        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:

  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 + 1}
          s_A(x1) = max{1, x1}
          app_A(x1,x2) = max{79, x1 + 54, x2 + 34}
          nil_A = 0
          cons_A(x1,x2) = 44
          sum_A(x1) = max{45, x1}
          plus_A(x1,x2) = max{x1 + 45, x2 + 15}
          |0|_A = 0
          pred_A(x1) = x1 + 31
    
      precedence:
      
        plus# > app = nil = cons = sum = plus > s = |0| = pred
      
      partial status:
    
        pi(plus#) = [1, 2]
        pi(s) = [1]
        pi(app) = [1]
        pi(nil) = []
        pi(cons) = []
        pi(sum) = [1]
        pi(plus) = [1, 2]
        pi(|0|) = []
        pi(pred) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = x2 + 1
          s_A(x1) = x1 + 43
          app_A(x1,x2) = x1 + 44
          nil_A = 40
          cons_A(x1,x2) = 49
          sum_A(x1) = max{80, x1 + 21}
          plus_A(x1,x2) = max{x1 + 10, x2 + 10}
          |0|_A = 39
          pred_A(x1) = max{26, x1 + 1}
    
      precedence:
      
        plus# = s = app = nil = cons = sum = plus = |0| = pred
      
      partial status:
    
        pi(plus#) = []
        pi(s) = [1]
        pi(app) = [1]
        pi(nil) = []
        pi(cons) = []
        pi(sum) = []
        pi(plus) = [2]
        pi(|0|) = []
        pi(pred) = [1]
    

The next rules are strictly ordered:

  p1

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