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, r9, r12

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1) = x1
              app_A(x1,x2) = x1 + x2 + 1
              cons_A(x1,x2) = 1
              sum_A(x1) = 1
              nil_A() = 1
              plus_A(x1,x2) = 2
              s_A(x1) = 1
              if_A(x1,x2,x3) = x2 + x3 + 1
              gt_A(x1,x2) = x2 + 1
              not_A(x1) = 1
              id_A(x1) = x1 + 1
              sum#!_A(x1) = 4
              app!_A(x1,x2) = 5
              cons!_A(x1,x2) = 3
              sum!_A(x1) = 3
              nil!_A() = 4
              plus!_A(x1,x2) = 4
              s!_A(x1) = 2
              if!_A(x1,x2,x3) = 1
              gt!_A(x1,x2) = 1
              not!_A(x1) = 3
              id!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            app_A(x1,x2) = max{x1 + 222, x2 + 55}
            cons_A(x1,x2) = 156
            sum_A(x1) = 168
            nil_A = 18
            plus_A(x1,x2) = max{279, x1 - 79, x2 - 16}
            s_A(x1) = 123
            if_A(x1,x2,x3) = max{77, x2 - 46, x3 - 15}
            gt_A(x1,x2) = max{x1 + 31, x2 + 154}
            not_A(x1) = 0
            id_A(x1) = max{93, x1 - 62}
            sum#!_A(x1) = max{15, x1 - 135}
            app!_A(x1,x2) = max{13, x1 + 3, x2 - 155}
            cons!_A(x1,x2) = max{2, x2 - 140}
            sum!_A(x1) = 14
            nil!_A = 15
            plus!_A(x1,x2) = max{17, x1 - 78}
            s!_A(x1) = 14
            if!_A(x1,x2,x3) = x2 + 62
            gt!_A(x1,x2) = max{11, x1 - 124}
            not!_A(x1) = max{221, x1 - 155}
            id!_A(x1) = max{14, x1 + 12}
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      pi(plus) = []
      pi(s) = []
      pi(if) = []
      pi(gt) = []
      pi(not) = [1]
      pi(id) = []

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              cons_A(x1,x2) = x1 + x2 + 8
              app#!_A(x1,x2) = x1 + 12
              cons!_A(x1,x2) = x2 + 11
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 294}
            cons_A(x1,x2) = max{x1 - 36, x2 + 293}
            app#!_A(x1,x2) = max{440, x2 + 1}
            cons!_A(x1,x2) = x1 + 439
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1) = 0
              cons_A(x1,x2) = x2 + 7
              plus_A(x1,x2) = 13
              sum#!_A(x1) = x1 + 11
              cons!_A(x1,x2) = x2 + 10
              plus!_A(x1,x2) = x1 + 26
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            cons_A(x1,x2) = max{374, x2 + 40}
            plus_A(x1,x2) = max{31, x1 - 74, x2 - 21}
            sum#!_A(x1) = max{4, x1 - 75}
            cons!_A(x1,x2) = x2 + 5
            plus!_A(x1,x2) = x2 + 92
        
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = [2]
      pi(plus) = []

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = 0
              s_A(x1) = 2
              if_A(x1,x2,x3) = x2 + x3 + 1
              gt_A(x1,x2) = 1
              not_A(x1) = 3
              id_A(x1) = x1 + 8
              true_A() = 1
              false_A() = 1
              zero_A() = 1
              plus#!_A(x1,x2) = 5
              s!_A(x1) = x1 + 6
              if!_A(x1,x2,x3) = 7
              gt!_A(x1,x2) = 8
              not!_A(x1) = x1 + 8
              id!_A(x1) = 0
              true!_A() = 7
              false!_A() = 0
              zero!_A() = 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{73, x1 + 52}
            if_A(x1,x2,x3) = max{65, x2 + 16, x3}
            gt_A(x1,x2) = max{x1 - 165, x2 + 99}
            not_A(x1) = x1 + 181
            id_A(x1) = max{47, x1 + 15}
            true_A = 181
            false_A = 0
            zero_A = 90
            plus#!_A(x1,x2) = max{2, x1 - 46, x2 - 98}
            s!_A(x1) = 0
            if!_A(x1,x2,x3) = max{162, x2 - 8, x3 - 19}
            gt!_A(x1,x2) = max{69, x1 - 19, x2 - 97}
            not!_A(x1) = x1 + 161
            id!_A(x1) = 0
            true!_A = 68
            false!_A = 70
            zero!_A = 67
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(gt) = [1]
      pi(not) = [1]
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p3



-- SCC decomposition.

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)))

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, p2}


-- 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)))

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, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = 0
              s_A(x1) = 4
              if_A(x1,x2,x3) = x2 + x3 + 6
              gt_A(x1,x2) = 1
              not_A(x1) = x1 + 8
              id_A(x1) = x1
              true_A() = 1
              false_A() = 1
              zero_A() = 1
              plus#!_A(x1,x2) = 10
              s!_A(x1) = x1 + 18
              if!_A(x1,x2,x3) = 12
              gt!_A(x1,x2) = 11
              not!_A(x1) = x1 + 13
              id!_A(x1) = 23
              true!_A() = 19
              false!_A() = 0
              zero!_A() = 20
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{36, x1 - 87, x2 - 59}
            s_A(x1) = max{78, x1 + 30}
            if_A(x1,x2,x3) = max{32, x2, x3}
            gt_A(x1,x2) = max{89, x2 + 88}
            not_A(x1) = max{59, x1 - 30}
            id_A(x1) = max{95, x1}
            true_A = 168
            false_A = 89
            zero_A = 154
            plus#!_A(x1,x2) = max{97, x1 + 30, x2 + 1}
            s!_A(x1) = 0
            if!_A(x1,x2,x3) = max{126, x2 - 15, x3 - 96}
            gt!_A(x1,x2) = max{32, x1 - 155, x2 - 79}
            not!_A(x1) = x1 + 73
            id!_A(x1) = max{0, x1 - 14}
            true!_A = 74
            false!_A = 0
            zero!_A = 75
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(gt) = [1, 2]
      pi(not) = []
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p1



-- 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

  r4, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x1
              id_A(x1) = x1 + 1
              s_A(x1) = x1 + 1
              if_A(x1,x2,x3) = x2 + x3 + 5
              gt_A(x1,x2) = x1 + 1
              sum_A(x1) = x1 + 1
              cons_A(x1,x2) = x1 + x2 + 1
              nil_A() = 0
              true_A() = 1
              false_A() = 1
              zero_A() = 1
              plus#!_A(x1,x2) = x1
              id!_A(x1) = 3
              s!_A(x1) = 8
              if!_A(x1,x2,x3) = 7
              gt!_A(x1,x2) = x2 + 2
              sum!_A(x1) = 12
              cons!_A(x1,x2) = 9
              nil!_A() = 0
              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 - 46, x2 - 73}
            id_A(x1) = max{6, x1 + 2}
            s_A(x1) = max{50, x1 + 5}
            if_A(x1,x2,x3) = max{41, x1 - 1, x2 + 4, x3}
            gt_A(x1,x2) = max{73, x1 - 15, x2 - 6}
            sum_A(x1) = max{575, x1 + 169}
            cons_A(x1,x2) = max{20, x1 - 334, x2 - 715}
            nil_A = 98
            true_A = 36
            false_A = 0
            zero_A = 0
            plus#!_A(x1,x2) = max{31, x1 - 3}
            id!_A(x1) = x1 + 7
            s!_A(x1) = x1 + 263
            if!_A(x1,x2,x3) = max{87, x1 - 3, x2 - 38, x3 - 20}
            gt!_A(x1,x2) = max{264, x1 - 18, x2 - 51}
            sum!_A(x1) = x1 + 261
            cons!_A(x1,x2) = max{0, x1 - 407, x2 - 99}
            nil!_A = 0
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = []
      pi(id) = [1]
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(gt) = [2]
      pi(sum) = [1]
      pi(cons) = [1]
      pi(nil) = []
      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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              gt#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 3
              gt#!_A(x1,x2) = x1 + x2
              s!_A(x1) = x1 + 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 1, x2 - 1}
            s_A(x1) = max{8, x1}
            gt#!_A(x1,x2) = max{23, x1 - 9, x2 - 9}
            s!_A(x1) = x1 + 22
        
  
    partial status:
  
      pi(gt#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.