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, r15, r16

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = max{3, x1}
        app_A(x1,x2) = max{101, x1 + 50, x2 + 98}
        cons_A(x1,x2) = 2
        sum_A(x1) = 2
        nil_A = 2
        plus_A(x1,x2) = max{x1 + 78, x2 + 48}
        gt_A(x1,x2) = max{3, x1 - 27}
        s_A(x1) = x1 + 28
        zero_A = 95
        true_A = 2
        false_A = 68
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1) = x1 + 9
          app_A(x1,x2) = max{12, x1 + 9, x2 + 7}
          cons_A(x1,x2) = max{10, x2}
          sum_A(x1) = max{10, x1 - 18}
          nil_A = 1
          plus_A(x1,x2) = 16
          gt_A(x1,x2) = 0
          s_A(x1) = 63
          zero_A = 0
          true_A = 0
          false_A = 0
          sum#!_A(x1) = 20
          app!_A(x1,x2) = 21
          cons!_A(x1,x2) = x2 + 10
          sum!_A(x1) = 12
          nil!_A = 11
          plus!_A(x1,x2) = 0
          gt!_A(x1,x2) = max{95, x1 - 64, x2 + 32}
          s!_A(x1) = x1 + 31
          zero!_A = 34
          true!_A = 33
          false!_A = 33
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      pi(plus) = []
      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: 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x2 + 183
        cons_A(x1,x2) = max{181, x1 - 359, x2 - 421}
  
    base order 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{8, x1 - 39, x2 + 5}
          app#!_A(x1,x2) = max{1, x1 - 6}
          cons!_A(x1,x2) = max{0, x2 - 710}
  
    partial status:
  
      pi(app#) = []
      pi(cons) = []

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: 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = max{67, x1 + 11}
        cons_A(x1,x2) = max{x1 + 247, x2 + 33}
        plus_A(x1,x2) = max{21, x2 + 7}
        s_A(x1) = 7
        if_A(x1,x2,x3) = max{12, x1 - 14, x2 + 5, x3 + 7}
        gt_A(x1,x2) = 15
        not_A(x1) = max{13, x1 - 7}
        id_A(x1) = max{11, x1 + 2}
        zero_A = 2
        true_A = 1
        false_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1) = 0
          cons_A(x1,x2) = max{52, x1 - 1, x2 - 15}
          plus_A(x1,x2) = max{64, x1 + 8, x2 + 8}
          s_A(x1) = max{29, x1}
          if_A(x1,x2,x3) = max{28, x1 - 9, x2, x3}
          gt_A(x1,x2) = max{36, x2 - 24}
          not_A(x1) = x1 + 29
          id_A(x1) = max{56, x1}
          zero_A = 1
          true_A = 29
          false_A = 29
          sum#!_A(x1) = x1 + 32
          cons!_A(x1,x2) = max{0, x2 - 57}
          plus!_A(x1,x2) = max{0, x2 - 59}
          s!_A(x1) = max{0, x1 - 64}
          if!_A(x1,x2,x3) = max{2, x1 - 29, x2 - 59, x3 - 60}
          gt!_A(x1,x2) = max{0, x1 - 59, x2 - 59}
          not!_A(x1) = max{1, x1 - 37}
          id!_A(x1) = max{0, x1 - 60}
          zero!_A = 1
          true!_A = 0
          false!_A = 0
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [2]
      pi(plus) = []
      pi(s) = []
      pi(if) = []
      pi(gt) = []
      pi(not) = []
      pi(id) = []
      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: 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{57, x1 + 2, x2 - 8}
        s_A(x1) = max{83, x1 + 13}
        if_A(x1,x2,x3) = max{x2 + 6, x3}
        gt_A(x1,x2) = max{17, x1 - 66, x2 - 84}
        not_A(x1) = 17
        id_A(x1) = max{48, x1 + 1}
        true_A = 16
        false_A = 1
        zero_A = 65
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 19, x2 - 19}
          s_A(x1) = x1 + 17
          if_A(x1,x2,x3) = max{x1 - 17, x2 + 17, x3}
          gt_A(x1,x2) = max{19, x1 - 35, x2 - 18}
          not_A(x1) = max{34, x1 - 2}
          id_A(x1) = max{1, x1}
          true_A = 1
          false_A = 0
          zero_A = 0
          plus#!_A(x1,x2) = 0
          s!_A(x1) = 0
          if!_A(x1,x2,x3) = max{53, x1 + 17, x2 + 4}
          gt!_A(x1,x2) = max{4, x1 - 22, x2 - 22}
          not!_A(x1) = max{4, x1 - 1}
          id!_A(x1) = 0
          true!_A = 0
          false!_A = 0
          zero!_A = 1
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(if) = [3]
      pi(gt) = []
      pi(not) = []
      pi(id) = []
      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: 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:

  {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: 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

  r12, r13, r15, r16, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = 1
        id_A(x1) = max{2, x1 - 883}
        s_A(x1) = x1 + 112
        if_A(x1,x2,x3) = max{162, x1 - 164, x2 + 56, x3 + 56}
        gt_A(x1,x2) = max{102, x1 - 388, x2 - 275}
        true_A = 0
        false_A = 0
        zero_A = 276
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 69, x2 - 7}
          id_A(x1) = max{15, x1 + 3}
          s_A(x1) = x1 + 6
          if_A(x1,x2,x3) = max{x1 - 17, x2 + 3, x3}
          gt_A(x1,x2) = max{20, x1 - 32, x2 - 7}
          true_A = 20
          false_A = 20
          zero_A = 0
          plus#!_A(x1,x2) = x1 + 2
          id!_A(x1) = 8
          s!_A(x1) = 18
          if!_A(x1,x2,x3) = max{19, x1 - 22, x2 - 40, x3 + 4}
          gt!_A(x1,x2) = max{2, x1 - 55, x2 - 55}
          true!_A = 1
          false!_A = 0
          zero!_A = 2
  
    partial status:
  
      pi(plus#) = []
      pi(id) = []
      pi(s) = []
      pi(if) = [2]
      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: 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        gt#_A(x1,x2) = 0
        s_A(x1) = max{1, x1 - 26}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          gt#_A(x1,x2) = max{0, x1 - 248, x2 - 248}
          s_A(x1) = max{266, x1 + 247}
          gt#!_A(x1,x2) = max{4, x2 - 242}
          s!_A(x1) = max{3, x1 - 686}
  
    partial status:
  
      pi(gt#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.