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, r11, r12, r13, r15, r16, r17

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 - 12}
            app_A(x1,x2) = max{17, x1 + 13, x2 + 2}
            cons_A(x1,x2) = max{12, x2}
            sum_A(x1) = 13
            nil_A = 2
            plus_A(x1,x2) = max{119, x1, x2}
            id_A(x1) = max{117, x1}
            if_A(x1,x2,x3) = max{x1 - 144, x2, x3 + 2}
            true_A = 4
            false_A = 4
            gt_A(x1,x2) = max{4, x1 - 66, x2 - 19}
            s_A(x1) = max{47, x1}
            zero_A = 0
            sum#!_A(x1) = max{1, x1 - 16}
            app!_A(x1,x2) = max{37, x1 + 25, x2 + 13}
            cons!_A(x1,x2) = max{17, x2 + 7}
            sum!_A(x1) = max{18, x1 - 1}
            nil!_A = 16
            plus!_A(x1,x2) = max{x1 + 21, x2 - 25}
            id!_A(x1) = max{0, x1 - 25}
            if!_A(x1,x2,x3) = max{142, x2 + 138, x3 - 48}
            true!_A = 140
            false!_A = 66
            gt!_A(x1,x2) = max{257, x1 - 48, x2 - 48}
            s!_A(x1) = max{257, x1 + 139}
            zero!_A = 141
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 46}
            app_A(x1,x2) = max{42, x1 + 1, x2 + 21}
            cons_A(x1,x2) = max{22, x2}
            sum_A(x1) = max{23, x1 - 45}
            nil_A = 23
            plus_A(x1,x2) = max{21, x1, x2}
            id_A(x1) = max{9, x1}
            if_A(x1,x2,x3) = max{22, x1 - 10, x2, x3}
            true_A = 9
            false_A = 1
            gt_A(x1,x2) = max{9, x1 - 24, x2 - 24}
            s_A(x1) = max{13, x1}
            zero_A = 0
            sum#!_A(x1) = 1
            app!_A(x1,x2) = max{2, x1 - 24, x2 - 24}
            cons!_A(x1,x2) = max{0, x2 - 70}
            sum!_A(x1) = max{0, x1 - 45}
            nil!_A = 0
            plus!_A(x1,x2) = max{5, x1 - 25}
            id!_A(x1) = max{1, x1 - 36}
            if!_A(x1,x2,x3) = max{1, x1 - 2, x2 - 2, x3 - 28}
            true!_A = 0
            false!_A = 0
            gt!_A(x1,x2) = max{4, x1 - 24, x2 - 26}
            s!_A(x1) = max{6, x1 - 14}
            zero!_A = 1
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      pi(plus) = [1]
      pi(id) = [1]
      pi(if) = [1, 2, 3]
      pi(true) = []
      pi(false) = []
      pi(gt) = [1, 2]
      pi(s) = [1]
      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: 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:
  
      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 - 8}
            cons_A(x1,x2) = max{12, x1 - 16, x2 + 7}
            app#!_A(x1,x2) = max{3, x1 - 8}
            cons!_A(x1,x2) = x2 + 579
        
        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{10, x1 - 16, x2 + 5}
            app#!_A(x1,x2) = max{19, x1 - 11, x2 - 1}
            cons!_A(x1,x2) = x2 + 18
        
  
    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

  r11, r12, r13, r15

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) = max{22, x2 + 14}
            plus_A(x1,x2) = max{1, x1 - 72}
            id_A(x1) = max{131, x1 + 13}
            if_A(x1,x2,x3) = max{147, x2, x3 + 1}
            true_A = 1
            false_A = 40
            gt_A(x1,x2) = 39
            s_A(x1) = 74
            zero_A = 34
            sum#!_A(x1) = max{1, x1 - 20}
            cons!_A(x1,x2) = max{12, x2 - 11}
            plus!_A(x1,x2) = max{23, x1 - 124, x2 - 110}
            id!_A(x1) = max{38, x1 - 109}
            if!_A(x1,x2,x3) = max{96, x2 - 168, x3 - 35}
            true!_A = 34
            false!_A = 1
            gt!_A(x1,x2) = max{98, x1 - 35, x2 - 35}
            s!_A(x1) = max{97, x1 + 36}
            zero!_A = 35
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = x1
            cons_A(x1,x2) = x2
            plus_A(x1,x2) = max{65, x2 + 4}
            id_A(x1) = max{24, x1 + 17}
            if_A(x1,x2,x3) = max{35, x2 + 18, x3 + 26}
            true_A = 6
            false_A = 19
            gt_A(x1,x2) = max{x1 + 19, x2}
            s_A(x1) = 16
            zero_A = 0
            sum#!_A(x1) = max{5, x1 - 75}
            cons!_A(x1,x2) = max{6, x2 + 4}
            plus!_A(x1,x2) = 0
            id!_A(x1) = max{1, x1 - 10}
            if!_A(x1,x2,x3) = max{22, x2 - 25, x3 - 25}
            true!_A = 0
            false!_A = 20
            gt!_A(x1,x2) = max{35, x1 - 17, x2 + 19}
            s!_A(x1) = 0
            zero!_A = 21
        
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = []
      pi(plus) = [1]
      pi(id) = [1]
      pi(if) = [2, 3]
      pi(true) = []
      pi(false) = []
      pi(gt) = [2]
      pi(s) = [1]
      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: 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:
  
      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 - 64, x2 - 64}
            s_A(x1) = x1 + 40
            if_A(x1,x2,x3) = max{17, x1 - 78, x2 + 11, x3}
            gt_A(x1,x2) = max{79, x1 - 41, x2 - 41}
            not_A(x1) = max{141, x1 - 78}
            id_A(x1) = max{3, x1}
            true_A = 1
            false_A = 3
            zero_A = 0
            plus#!_A(x1,x2) = max{115, x1 + 100, x2 + 76}
            s!_A(x1) = x1 + 86
            if!_A(x1,x2,x3) = max{2, x1 - 215, x2 - 4, x3 - 41}
            gt!_A(x1,x2) = max{141, x1 - 257, x2 - 255}
            not!_A(x1) = max{141, x1 - 80}
            id!_A(x1) = max{1, x1 - 535}
            true!_A = 4
            false!_A = 0
            zero!_A = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 844, x2 - 844}
            s_A(x1) = max{313, x1 + 261}
            if_A(x1,x2,x3) = max{265, x1 - 78, x2 + 259, x3}
            gt_A(x1,x2) = max{79, x1 - 314, x2 - 290}
            not_A(x1) = max{921, x1 - 78}
            id_A(x1) = max{286, x1 + 1}
            true_A = 79
            false_A = 79
            zero_A = 0
            plus#!_A(x1,x2) = max{16, x2 - 844}
            s!_A(x1) = x1 + 839
            if!_A(x1,x2,x3) = max{15, x1 - 1132, x2 - 1131, x3 - 1417}
            gt!_A(x1,x2) = max{840, x1 - 526, x2 - 314}
            not!_A(x1) = max{840, x1 - 292}
            id!_A(x1) = max{14, x1 - 1052}
            true!_A = 18
            false!_A = 0
            zero!_A = 19
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [1, 2, 3]
      pi(gt) = [1]
      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: 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:
  
      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 - 414, x2 - 297}
            id_A(x1) = max{116, x1 + 15}
            s_A(x1) = x1 + 59
            if_A(x1,x2,x3) = max{237, x1 + 60, x2 + 59, x3}
            gt_A(x1,x2) = max{236, x1 - 72, x2 - 60}
            true_A = 189
            false_A = 236
            zero_A = 0
            plus#!_A(x1,x2) = max{59, x1 - 355, x2 - 238}
            id!_A(x1) = max{5, x1 - 1034}
            s!_A(x1) = max{4, x1 - 618}
            if!_A(x1,x2,x3) = max{60, x1 - 414, x2 - 708, x3 - 190}
            gt!_A(x1,x2) = max{6, x1 - 190, x2 - 60}
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 65, x2 - 44}
            id_A(x1) = max{18, x1}
            s_A(x1) = max{4, x1 + 3}
            if_A(x1,x2,x3) = max{43, x1 - 1, x2 + 3, x3}
            gt_A(x1,x2) = max{26, x1 - 5, x2 - 4}
            true_A = 26
            false_A = 23
            zero_A = 0
            plus#!_A(x1,x2) = max{4, x1 - 25, x2 - 60}
            id!_A(x1) = max{1, x1 - 126}
            s!_A(x1) = max{3, x1 - 66}
            if!_A(x1,x2,x3) = max{5, x1 - 52, x2 - 27, x3 - 27}
            gt!_A(x1,x2) = max{5, x1 - 67, x2 - 5}
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = [1]
      pi(s) = [1]
      pi(if) = [1, 2, 3]
      pi(gt) = [2]
      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:
  
      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 - 189, x2 - 189}
            s_A(x1) = x1 + 188
            gt#!_A(x1,x2) = max{6, x1 - 189, x2 - 189}
            s!_A(x1) = max{5, x1 + 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 34, x2 - 34}
            s_A(x1) = max{39, x1 + 33}
            gt#!_A(x1,x2) = max{3, x1 - 150, x2 - 40}
            s!_A(x1) = max{2, x1 - 151}
        
  
    partial status:
  
      pi(gt#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.