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)))))
  sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
  pred(cons(s(x),nil())) -> cons(x,nil())
  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: sum#(plus(cons(|0|(),x),cons(y,l))) -> pred#(sum(cons(s(x),cons(y,l))))
p8: sum#(plus(cons(|0|(),x),cons(y,l))) -> sum#(cons(s(x),cons(y,l)))
p9: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p10: plus#(s(x),s(y)) -> if#(gt(x,y),x,y)
p11: plus#(s(x),s(y)) -> gt#(x,y)
p12: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p13: plus#(s(x),s(y)) -> not#(gt(x,y))
p14: plus#(s(x),s(y)) -> id#(x)
p15: plus#(s(x),s(y)) -> id#(y)
p16: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))
p17: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p18: plus#(s(x),x) -> gt#(x,x)
p19: plus#(s(x),x) -> id#(x)
p20: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p21: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p22: plus#(id(x),s(y)) -> gt#(s(y),y)
p23: not#(x) -> if#(x,false(),true())
p24: 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: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r8: pred(cons(s(x),nil())) -> cons(x,nil())
r9: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r10: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r11: plus(zero(),y) -> y
r12: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r13: id(x) -> x
r14: if(true(),x,y) -> x
r15: if(false(),x,y) -> y
r16: not(x) -> if(x,false(),true())
r17: gt(s(x),zero()) -> true()
r18: gt(zero(),y) -> false()
r19: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1}
  {p2}
  {p9, p16, p20}
  {p24}


-- 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: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r8: pred(cons(s(x),nil())) -> cons(x,nil())
r9: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r10: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r11: plus(zero(),y) -> y
r12: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r13: id(x) -> x
r14: if(true(),x,y) -> x
r15: if(false(),x,y) -> y
r16: not(x) -> if(x,false(),true())
r17: gt(s(x),zero()) -> true()
r18: gt(zero(),y) -> false()
r19: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r17, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            app_A(x1,x2) = max{x1 + 12, x2 + 7}
            cons_A(x1,x2) = max{4, x2}
            sum_A(x1) = 4
            nil_A = 2
            plus_A(x1,x2) = max{25, x1 + 10}
            gt_A(x1,x2) = max{x1 - 27, x2}
            s_A(x1) = max{24, x1}
            zero_A = 28
            true_A = 10
            false_A = 1
            sum#!_A(x1) = max{18, x1 + 13}
            app!_A(x1,x2) = max{x1 + 16, x2 + 13}
            cons!_A(x1,x2) = x2 + 3
            sum!_A(x1) = 6
            nil!_A = 1
            plus!_A(x1,x2) = max{x1 + 35, x2 + 9}
            gt!_A(x1,x2) = 51
            s!_A(x1) = 52
            zero!_A = 2
            true!_A = 1
            false!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            app_A(x1,x2) = max{x1 + 36, x2}
            cons_A(x1,x2) = 53
            sum_A(x1) = 53
            nil_A = 44
            plus_A(x1,x2) = 36
            gt_A(x1,x2) = max{x1 - 12, x2 + 3}
            s_A(x1) = 20
            zero_A = 38
            true_A = 41
            false_A = 26
            sum#!_A(x1) = max{71, x1 + 36}
            app!_A(x1,x2) = x1 + 8
            cons!_A(x1,x2) = max{45, x2 + 2}
            sum!_A(x1) = 46
            nil!_A = 47
            plus!_A(x1,x2) = 55
            gt!_A(x1,x2) = 56
            s!_A(x1) = 57
            zero!_A = 2
            true!_A = 1
            false!_A = 1
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      pi(plus) = [1]
      pi(gt) = []
      pi(s) = []
      pi(zero) = []
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p1

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: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r8: pred(cons(s(x),nil())) -> cons(x,nil())
r9: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r10: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r11: plus(zero(),y) -> y
r12: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r13: id(x) -> x
r14: if(true(),x,y) -> x
r15: if(false(),x,y) -> y
r16: not(x) -> if(x,false(),true())
r17: gt(s(x),zero()) -> true()
r18: gt(zero(),y) -> false()
r19: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 6}
            cons_A(x1,x2) = max{94, x1 - 15, x2 + 5}
            app#!_A(x1,x2) = max{1, x1 - 95, x2 - 1}
            cons!_A(x1,x2) = max{0, x1 - 399, x2 - 520}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 6}
            cons_A(x1,x2) = max{34, x1 - 12, x2 + 5}
            app#!_A(x1,x2) = max{1, x1 - 1381, x2 - 1}
            cons!_A(x1,x2) = max{5, x1 - 614, x2 - 1382}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [1, 2]

The following rules are removed:

  p1

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: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r8: pred(cons(s(x),nil())) -> cons(x,nil())
r9: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r10: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r11: plus(zero(),y) -> y
r12: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r13: id(x) -> x
r14: if(true(),x,y) -> x
r15: if(false(),x,y) -> y
r16: not(x) -> if(x,false(),true())
r17: gt(s(x),zero()) -> true()
r18: gt(zero(),y) -> false()
r19: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 37}
            cons_A(x1,x2) = x2 + 18
            plus_A(x1,x2) = max{0, x2 - 24}
            sum#!_A(x1) = 3
            cons!_A(x1,x2) = max{0, x2 - 22}
            plus!_A(x1,x2) = x1 + 43
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 11}
            cons_A(x1,x2) = x2 + 21
            plus_A(x1,x2) = 83
            sum#!_A(x1) = max{5, x1 - 119}
            cons!_A(x1,x2) = max{7, x2 - 2}
            plus!_A(x1,x2) = max{x1 + 57, x2 + 38}
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [2]
      pi(plus) = [1, 2]

The following rules are removed:

  p1

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: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r8: pred(cons(s(x),nil())) -> cons(x,nil())
r9: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r10: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r11: plus(zero(),y) -> y
r12: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r13: id(x) -> x
r14: if(true(),x,y) -> x
r15: if(false(),x,y) -> y
r16: not(x) -> if(x,false(),true())
r17: gt(s(x),zero()) -> true()
r18: gt(zero(),y) -> false()
r19: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r13, r14, r15, r16, r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{5, x1 - 98, x2 - 161}
            s_A(x1) = max{166, x1 + 97}
            if_A(x1,x2,x3) = max{35, x1 - 264, x2, x3}
            gt_A(x1,x2) = max{245, x1 - 167, x2 - 174}
            not_A(x1) = max{262, x1 - 9}
            id_A(x1) = max{162, x1 + 86}
            true_A = 1
            false_A = 245
            zero_A = 0
            plus#!_A(x1,x2) = max{x1 + 6, x2 + 1}
            s!_A(x1) = 0
            if!_A(x1,x2,x3) = max{104, x1 - 139, x2 + 27, x3 - 167}
            gt!_A(x1,x2) = max{166, x1 - 151, x2 - 167}
            not!_A(x1) = max{273, x1 + 271}
            id!_A(x1) = x1 + 105
            true!_A = 0
            false!_A = 165
            zero!_A = 166
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 162, x2 - 162}
            s_A(x1) = max{53, x1 + 50}
            if_A(x1,x2,x3) = max{64, x1 - 12, x2 + 17, x3}
            gt_A(x1,x2) = max{173, x1 - 54, x2 - 55}
            not_A(x1) = max{64, x1 - 10}
            id_A(x1) = max{144, x1}
            true_A = 50
            false_A = 0
            zero_A = 53
            plus#!_A(x1,x2) = 262
            s!_A(x1) = max{8, x1 - 84}
            if!_A(x1,x2,x3) = max{96, x1 - 65, x2 - 145}
            gt!_A(x1,x2) = max{5, x1 - 54, x2 - 54}
            not!_A(x1) = max{263, x1 + 88}
            id!_A(x1) = 6
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [1, 2, 3]
      pi(gt) = [2]
      pi(not) = [1]
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

p1: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(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: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r8: pred(cons(s(x),nil())) -> cons(x,nil())
r9: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r10: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r11: plus(zero(),y) -> y
r12: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r13: id(x) -> x
r14: if(true(),x,y) -> x
r15: if(false(),x,y) -> y
r16: not(x) -> if(x,false(),true())
r17: gt(s(x),zero()) -> true()
r18: gt(zero(),y) -> false()
r19: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(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: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r8: pred(cons(s(x),nil())) -> cons(x,nil())
r9: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r10: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r11: plus(zero(),y) -> y
r12: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r13: id(x) -> x
r14: if(true(),x,y) -> x
r15: if(false(),x,y) -> y
r16: not(x) -> if(x,false(),true())
r17: gt(s(x),zero()) -> true()
r18: gt(zero(),y) -> false()
r19: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r14, r15, r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 791, x2 - 386}
            id_A(x1) = x1 + 582
            s_A(x1) = max{211, x1 + 188}
            if_A(x1,x2,x3) = max{x1 + 377, x2, x3}
            gt_A(x1,x2) = max{8, x1 - 410, x2 - 189}
            true_A = 8
            false_A = 8
            zero_A = 0
            plus#!_A(x1,x2) = max{581, x1 - 1, x2 - 525}
            id!_A(x1) = max{4, x1 - 1135}
            s!_A(x1) = max{3, x1 - 371}
            if!_A(x1,x2,x3) = max{582, x1 - 937, x2 - 583, x3 - 946}
            gt!_A(x1,x2) = max{583, x1 - 370, x2 - 946}
            true!_A = 1
            false!_A = 1
            zero!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 342, x2 - 180}
            id_A(x1) = max{159, x1 + 18}
            s_A(x1) = max{140, x1 + 121}
            if_A(x1,x2,x3) = max{178, x1 - 1, x2 + 121, x3}
            gt_A(x1,x2) = max{180, x1 - 141, x2 - 122}
            true_A = 180
            false_A = 180
            zero_A = 0
            plus#!_A(x1,x2) = max{5, x1 - 425, x2 - 263}
            id!_A(x1) = max{4, x1 - 426}
            s!_A(x1) = max{2, x1 - 804}
            if!_A(x1,x2,x3) = max{4, x1 - 527, x2 - 426, x3 - 463}
            gt!_A(x1,x2) = max{6, x1 - 262, x2 - 141}
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = [1]
      pi(s) = []
      pi(if) = [1, 2, 3]
      pi(gt) = []
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p1

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: sum(plus(cons(|0|(),x),cons(y,l))) -> pred(sum(cons(s(x),cons(y,l))))
r8: pred(cons(s(x),nil())) -> cons(x,nil())
r9: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r10: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r11: plus(zero(),y) -> y
r12: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r13: id(x) -> x
r14: if(true(),x,y) -> x
r15: if(false(),x,y) -> y
r16: not(x) -> if(x,false(),true())
r17: gt(s(x),zero()) -> true()
r18: gt(zero(),y) -> false()
r19: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 36, x2 - 36}
            s_A(x1) = x1 + 35
            gt#!_A(x1,x2) = max{6, x1 - 210, x2 - 36}
            s!_A(x1) = max{5, x1 - 211}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 82, x2 - 82}
            s_A(x1) = max{113, x1 + 81}
            gt#!_A(x1,x2) = max{28, x1 - 114, x2 - 114}
            s!_A(x1) = max{27, x1 - 223}
        
  
    partial status:
  
      pi(gt#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.