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(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
  plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
  plus(zero(),y) -> y
  plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
  id(x) -> x
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  not(x) -> if(x,false(),true())
  gt(s(x),zero()) -> true()
  gt(zero(),y) -> false()
  gt(s(x),s(y)) -> gt(x,y)

-- 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),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p8: plus#(s(x),s(y)) -> if#(gt(x,y),x,y)
p9: plus#(s(x),s(y)) -> gt#(x,y)
p10: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p11: plus#(s(x),s(y)) -> not#(gt(x,y))
p12: plus#(s(x),s(y)) -> id#(x)
p13: plus#(s(x),s(y)) -> id#(y)
p14: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))
p15: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p16: plus#(s(x),x) -> gt#(x,x)
p17: plus#(s(x),x) -> id#(x)
p18: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p19: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p20: plus#(id(x),s(y)) -> gt#(s(y),y)
p21: not#(x) -> if#(x,false(),true())
p22: gt#(s(x),s(y)) -> gt#(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(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1}
  {p2}
  {p7, p14, p18}
  {p22}


-- 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(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, 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:
            sum#_A(x1) = ((1,0),(1,1)) x1 + (2,1)
            app_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,7)
            cons_A(x1,x2) = x2 + (4,3)
            sum_A(x1) = (5,5)
            id_A(x1) = x1 + (4,4)
            if_A(x1,x2,x3) = x2 + x3 + (5,4)
            true_A() = (0,4)
            false_A() = (1,6)
            not_A(x1) = (7,15)
            gt_A(x1,x2) = (2,3)
            s_A(x1) = (0,1)
            zero_A() = (2,5)
            nil_A() = (1,1)
            plus_A(x1,x2) = ((1,0),(0,0)) x2 + (3,2)
    
      precedence:
      
        sum = true = false = gt = s = plus > sum# = app = cons = not = zero = nil > id > if
      
      partial status:
    
        pi(sum#) = [1]
        pi(app) = [1]
        pi(cons) = []
        pi(sum) = []
        pi(id) = [1]
        pi(if) = [2, 3]
        pi(true) = []
        pi(false) = []
        pi(not) = []
        pi(gt) = []
        pi(s) = []
        pi(zero) = []
        pi(nil) = []
        pi(plus) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sum#_A(x1) = ((0,0),(1,0)) x1 + (1,0)
            app_A(x1,x2) = x1 + (3,4)
            cons_A(x1,x2) = (2,4)
            sum_A(x1) = (2,4)
            id_A(x1) = x1 + (8,1)
            if_A(x1,x2,x3) = x2 + x3 + (3,2)
            true_A() = (1,1)
            false_A() = (1,1)
            not_A(x1) = (6,0)
            gt_A(x1,x2) = (0,0)
            s_A(x1) = (7,0)
            zero_A() = (2,2)
            nil_A() = (2,4)
            plus_A(x1,x2) = (9,0)
    
      precedence:
      
        sum = not = plus > if > sum# = app = cons = id = true = false = gt = s = zero = nil
      
      partial status:
    
        pi(sum#) = []
        pi(app) = []
        pi(cons) = []
        pi(sum) = []
        pi(id) = [1]
        pi(if) = [2, 3]
        pi(true) = []
        pi(false) = []
        pi(not) = []
        pi(gt) = []
        pi(s) = []
        pi(zero) = []
        pi(nil) = []
        pi(plus) = []
    

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(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

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(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  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:
        
          sum#_A(x1) = max{79, x1 + 18}
          cons_A(x1,x2) = max{47, x2 + 31}
          plus_A(x1,x2) = max{x1 + 79, x2 + 83}
          id_A(x1) = max{2, x1 + 1}
          if_A(x1,x2,x3) = max{3, x1 - 89, x2 + 1, x3 + 1}
          true_A = 25
          false_A = 25
          not_A(x1) = max{84, x1 + 27}
          gt_A(x1,x2) = 73
          s_A(x1) = 0
          zero_A = 0
    
      precedence:
      
        not > gt > sum# = cons > plus > false = s > if = zero > true > id
      
      partial status:
    
        pi(sum#) = [1]
        pi(cons) = [2]
        pi(plus) = []
        pi(id) = [1]
        pi(if) = [2, 3]
        pi(true) = []
        pi(false) = []
        pi(not) = [1]
        pi(gt) = []
        pi(s) = []
        pi(zero) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sum#_A(x1) = 0
          cons_A(x1,x2) = x2 + 29
          plus_A(x1,x2) = 15
          id_A(x1) = max{23, x1}
          if_A(x1,x2,x3) = max{46, x2 + 13, x3 + 13}
          true_A = 44
          false_A = 33
          not_A(x1) = x1 + 45
          gt_A(x1,x2) = 20
          s_A(x1) = 19
          zero_A = 0
    
      precedence:
      
        sum# = cons = plus = id = if = true = false = not = gt = s = zero
      
      partial status:
    
        pi(sum#) = []
        pi(cons) = [2]
        pi(plus) = []
        pi(id) = [1]
        pi(if) = []
        pi(true) = []
        pi(false) = []
        pi(not) = []
        pi(gt) = []
        pi(s) = []
        pi(zero) = []
    

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),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p3: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))

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(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  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:
        
          plus#_A(x1,x2) = max{64, x1 + 41, x2 + 33}
          s_A(x1) = max{32, x1 + 21}
          if_A(x1,x2,x3) = max{x2 + 10, x3}
          gt_A(x1,x2) = max{40, x2 + 20}
          not_A(x1) = x1 + 27
          id_A(x1) = max{22, x1}
          true_A = 8
          false_A = 0
          zero_A = 9
    
      precedence:
      
        s = not > plus# = if = gt = id = true = false = zero
      
      partial status:
    
        pi(plus#) = [1, 2]
        pi(s) = [1]
        pi(if) = [2, 3]
        pi(gt) = [2]
        pi(not) = [1]
        pi(id) = [1]
        pi(true) = []
        pi(false) = []
        pi(zero) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{139, x1 + 106, x2 + 75}
          s_A(x1) = x1 + 59
          if_A(x1,x2,x3) = max{14, x2 + 2}
          gt_A(x1,x2) = x2 + 13
          not_A(x1) = x1 + 13
          id_A(x1) = x1 + 33
          true_A = 12
          false_A = 12
          zero_A = 0
    
      precedence:
      
        plus# = s = if = gt = not = id = true = false = zero
      
      partial status:
    
        pi(plus#) = [2]
        pi(s) = [1]
        pi(if) = []
        pi(gt) = [2]
        pi(not) = [1]
        pi(id) = []
        pi(true) = []
        pi(false) = []
        pi(zero) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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: gt#(s(x),s(y)) -> gt#(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(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

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:
        
          gt#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        gt# = s
      
      partial status:
    
        pi(gt#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          gt#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        gt# = s
      
      partial status:
    
        pi(gt#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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