YES
We show the termination of the TRS R:

  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  quot(|0|(),s(y)) -> |0|()
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  minus(minus(x,y),z) -> minus(x,plus(y,z))
  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: minus#(s(x),s(y)) -> minus#(x,y)
p2: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p3: quot#(s(x),s(y)) -> minus#(x,y)
p4: plus#(s(x),y) -> plus#(x,y)
p5: minus#(minus(x,y),z) -> minus#(x,plus(y,z))
p6: minus#(minus(x,y),z) -> plus#(y,z)
p7: app#(cons(x,l),k) -> app#(l,k)
p8: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p9: sum#(cons(x,cons(y,l))) -> plus#(x,y)
p10: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p11: sum#(app(l,cons(x,cons(y,k)))) -> app#(l,sum(cons(x,cons(y,k))))
p12: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p13: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p14: plus#(s(x),s(y)) -> if#(gt(x,y),x,y)
p15: plus#(s(x),s(y)) -> gt#(x,y)
p16: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p17: plus#(s(x),s(y)) -> not#(gt(x,y))
p18: plus#(s(x),s(y)) -> id#(x)
p19: plus#(s(x),s(y)) -> id#(y)
p20: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))
p21: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p22: plus#(s(x),x) -> gt#(x,x)
p23: plus#(s(x),x) -> id#(x)
p24: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p25: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p26: plus#(id(x),s(y)) -> gt#(s(y),y)
p27: not#(x) -> if#(x,false(),true())
p28: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p5}
  {p10}
  {p8}
  {p4, p13, p20, p24}
  {p7}
  {p28}


-- Reduction pair.

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

p1: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r1, r2, r7, r22, r23

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:
          
            quot#_A(x1,x2) = max{0, x2 - 49}
            s_A(x1) = max{48, x1 + 8}
            minus_A(x1,x2) = x1 + 68
            |0|_A = 11
            plus_A(x1,x2) = max{7, x1 - 26}
            gt_A(x1,x2) = max{92, x1 - 49}
            zero_A = 9
            true_A = 1
            false_A = 1
            quot#!_A(x1,x2) = 0
            s!_A(x1) = 49
            minus!_A(x1,x2) = 0
            |0|!_A = 51
            plus!_A(x1,x2) = max{48, x1 + 42}
            gt!_A(x1,x2) = max{91, x2 - 49}
            zero!_A = 50
            true!_A = 49
            false!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = x2 + 1
            s_A(x1) = x1 + 6
            minus_A(x1,x2) = max{5, x1}
            |0|_A = 11
            plus_A(x1,x2) = max{76, x1 - 21, x2 - 23}
            gt_A(x1,x2) = max{109, x1 - 10}
            zero_A = 9
            true_A = 1
            false_A = 1
            quot#!_A(x1,x2) = x1 + 6
            s!_A(x1) = x1 + 110
            minus!_A(x1,x2) = 0
            |0|!_A = 117
            plus!_A(x1,x2) = max{x1 + 203, x2}
            gt!_A(x1,x2) = max{x1 - 7, x2 + 210}
            zero!_A = 111
            true!_A = 110
            false!_A = 1
        
  
    partial status:
  
      pi(quot#) = []
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(plus) = [2]
      pi(gt) = [2]
      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: minus#(s(x),s(y)) -> minus#(x,y)
p2: minus#(minus(x,y),z) -> minus#(x,plus(y,z))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r1, r18, r19, r20, r22, r23

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:
          
            minus#_A(x1,x2) = max{0, x1 - 37}
            s_A(x1) = max{18, x1}
            minus_A(x1,x2) = max{x1 + 36, x2 - 37}
            plus_A(x1,x2) = max{x1 - 9, x2}
            |0|_A = 32
            id_A(x1) = max{10, x1 + 9}
            if_A(x1,x2,x3) = max{34, x2, x3}
            true_A = 9
            false_A = 1
            gt_A(x1,x2) = max{9, x1 - 19, x2 - 19}
            zero_A = 0
            minus#!_A(x1,x2) = max{1, x1 - 37}
            s!_A(x1) = max{36, x1 - 4}
            minus!_A(x1,x2) = max{0, x1 - 38}
            plus!_A(x1,x2) = max{35, x1 + 19, x2 + 1}
            |0|!_A = 36
            id!_A(x1) = max{0, x1 - 14}
            if!_A(x1,x2,x3) = max{41, x1 - 10, x2 - 11, x3 - 19}
            true!_A = 0
            false!_A = 39
            gt!_A(x1,x2) = max{38, x1 - 19, x2 - 19}
            zero!_A = 40
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 81}
            s_A(x1) = max{83, x1}
            minus_A(x1,x2) = max{x1 + 80, x2 - 82}
            plus_A(x1,x2) = max{81, x1, x2}
            |0|_A = 83
            id_A(x1) = max{80, x1}
            if_A(x1,x2,x3) = max{82, x1 - 87, x2, x3}
            true_A = 1
            false_A = 1
            gt_A(x1,x2) = max{1, x1 - 84, x2 - 84}
            zero_A = 0
            minus#!_A(x1,x2) = max{78, x1 - 1}
            s!_A(x1) = 0
            minus!_A(x1,x2) = max{257, x1 - 84}
            plus!_A(x1,x2) = max{77, x2 - 85}
            |0|!_A = 1
            id!_A(x1) = max{0, x1 - 174}
            if!_A(x1,x2,x3) = max{77, x1 - 87, x2 - 86, x3 - 2}
            true!_A = 0
            false!_A = 2
            gt!_A(x1,x2) = max{1, x1 - 84, x2 - 84}
            zero!_A = 3
        
  
    partial status:
  
      pi(minus#) = []
      pi(s) = []
      pi(minus) = [1, 2]
      pi(plus) = []
      pi(|0|) = []
      pi(id) = [1]
      pi(if) = [1, 2, 3]
      pi(true) = []
      pi(false) = []
      pi(gt) = [1, 2]
      pi(zero) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: 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: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: 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:
          
            minus#_A(x1,x2) = max{0, x1 - 60, x2 - 60}
            s_A(x1) = max{63, x1 + 59}
            minus#!_A(x1,x2) = max{1, x1 - 282, x2 - 283}
            s!_A(x1) = max{2, x1 - 283}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 35, x2 - 35}
            s_A(x1) = max{37, x1 + 34}
            minus#!_A(x1,x2) = max{4, x1 - 38, x2 - 234}
            s!_A(x1) = max{3, x1 - 235}
        
  
    partial status:
  
      pi(minus#) = [2]
      pi(s) = [1]

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#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r8, r9, r10, r11, r12, r23

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) = x1
            app_A(x1,x2) = max{x1 + 5, x2}
            cons_A(x1,x2) = x2
            sum_A(x1) = 0
            nil_A = 0
            plus_A(x1,x2) = 0
            gt_A(x1,x2) = max{x1, x2 + 2}
            zero_A = 0
            false_A = 2
            sum#!_A(x1) = max{8, x1 - 2}
            app!_A(x1,x2) = max{x1 + 6, x2 + 7}
            cons!_A(x1,x2) = 2
            sum!_A(x1) = 2
            nil!_A = 1
            plus!_A(x1,x2) = 15
            gt!_A(x1,x2) = max{20, x1 - 8, x2 + 18}
            zero!_A = 20
            false!_A = 19
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{6, x1 - 3}
            app_A(x1,x2) = max{6, x1 + 4, x2}
            cons_A(x1,x2) = max{9, x2}
            sum_A(x1) = max{9, x1 - 15}
            nil_A = 6
            plus_A(x1,x2) = max{x1 - 16, x2}
            gt_A(x1,x2) = 20
            zero_A = 0
            false_A = 15
            sum#!_A(x1) = x1 + 14
            app!_A(x1,x2) = max{x1 + 3, x2 - 6}
            cons!_A(x1,x2) = max{6, x2 + 3}
            sum!_A(x1) = max{4, x1 + 1}
            nil!_A = 10
            plus!_A(x1,x2) = max{0, x2 - 21}
            gt!_A(x1,x2) = max{6, x1 - 22, x2 - 22}
            zero!_A = 2
            false!_A = 0
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      pi(plus) = []
      pi(gt) = []
      pi(zero) = []
      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: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r5, r6, r14, r16, r17, r20

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
            cons_A(x1,x2) = max{30, x2 + 6}
            plus_A(x1,x2) = max{243, x2 + 28}
            |0|_A = 217
            s_A(x1) = 215
            if_A(x1,x2,x3) = max{x1 + 79, x3 + 78}
            gt_A(x1,x2) = max{2, x1 - 1, x2 - 1}
            not_A(x1) = max{3, x1}
            id_A(x1) = x1
            zero_A = 0
            false_A = 3
            sum#!_A(x1) = x1 + 16
            cons!_A(x1,x2) = max{46, x2 + 44}
            plus!_A(x1,x2) = max{134, x1 - 162, x2 - 82}
            |0|!_A = 212
            s!_A(x1) = 52
            if!_A(x1,x2,x3) = 4
            gt!_A(x1,x2) = max{54, x1 - 4, x2 + 1}
            not!_A(x1) = 3
            id!_A(x1) = 1
            zero!_A = 3
            false!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            cons_A(x1,x2) = max{32, x2 + 26}
            plus_A(x1,x2) = max{4, x2 + 1}
            |0|_A = 3
            s_A(x1) = 3
            if_A(x1,x2,x3) = max{26, x2 + 4, x3 + 3}
            gt_A(x1,x2) = max{2, x1 - 4, x2 - 4}
            not_A(x1) = 26
            id_A(x1) = max{14, x1 + 8}
            zero_A = 0
            false_A = 12
            sum#!_A(x1) = max{31, x1 - 27}
            cons!_A(x1,x2) = max{36, x2}
            plus!_A(x1,x2) = max{4, x2 + 1}
            |0|!_A = 1
            s!_A(x1) = 0
            if!_A(x1,x2,x3) = max{x1 + 10, x2 - 11, x3 - 15}
            gt!_A(x1,x2) = max{6, x1 - 4, x2 - 4}
            not!_A(x1) = 0
            id!_A(x1) = max{5, x1 + 2}
            zero!_A = 1
            false!_A = 0
        
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = [2]
      pi(plus) = [1, 2]
      pi(|0|) = []
      pi(s) = []
      pi(if) = [2, 3]
      pi(gt) = []
      pi(not) = [1]
      pi(id) = []
      pi(zero) = []
      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),y) -> plus#(x,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))
p4: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r18, r19, r20, r21, r22, r23, r24

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 - 260, x2 - 260}
            s_A(x1) = max{40, x1 + 37}
            id_A(x1) = max{154, x1}
            if_A(x1,x2,x3) = max{125, x1 - 1, x2, x3}
            gt_A(x1,x2) = max{124, x1 - 41, x2 - 44}
            not_A(x1) = max{260, x1 - 1}
            true_A = 124
            false_A = 124
            zero_A = 0
            plus#!_A(x1,x2) = max{71, x1 - 152, x2 - 189}
            s!_A(x1) = max{0, x1 - 135}
            id!_A(x1) = max{0, x1 - 263}
            if!_A(x1,x2,x3) = max{1, x1 - 261, x2 - 305, x3 - 459}
            gt!_A(x1,x2) = max{1, x1 - 260, x2 - 305}
            not!_A(x1) = max{5, x1 - 235}
            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 - 32, x2 - 34}
            s_A(x1) = x1 + 17
            id_A(x1) = max{6, x1}
            if_A(x1,x2,x3) = max{16, x1 - 24, x2, x3}
            gt_A(x1,x2) = max{1, x1 - 21, x2 - 18}
            not_A(x1) = max{57, x1 - 19}
            true_A = 1
            false_A = 1
            zero_A = 17
            plus#!_A(x1,x2) = max{x1 + 79, x2 + 62}
            s!_A(x1) = x1 + 140
            id!_A(x1) = max{58, x1 - 94}
            if!_A(x1,x2,x3) = max{139, x1 + 137, x2 - 7, x3 - 18}
            gt!_A(x1,x2) = max{60, x1 - 18}
            not!_A(x1) = max{141, x1 + 138}
            true!_A = 59
            false!_A = 1
            zero!_A = 60
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(id) = [1]
      pi(if) = [2, 3]
      pi(gt) = [2]
      pi(not) = []
      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),y) -> plus#(x,y)
p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p3: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: plus#(s(x),y) -> plus#(x,y)
p2: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p3: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r18, r19, r20, r21, r22, r23, r24

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 - 157, x2 - 142}
            s_A(x1) = x1 + 130
            if_A(x1,x2,x3) = max{116, x1 - 3, x2 + 18, x3}
            gt_A(x1,x2) = max{129, x1 - 131, x2 - 131}
            not_A(x1) = max{134, x1 - 3}
            id_A(x1) = max{123, x1}
            true_A = 1
            false_A = 3
            zero_A = 0
            plus#!_A(x1,x2) = max{2, x1 - 127, x2 - 140}
            s!_A(x1) = max{10, x1 - 685}
            if!_A(x1,x2,x3) = max{11, x1 - 550, x2 - 547, x3 - 131}
            gt!_A(x1,x2) = max{9, x1 - 686, x2 - 686}
            not!_A(x1) = max{12, x1 - 545}
            id!_A(x1) = 4
            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 - 134, x2 - 130}
            s_A(x1) = x1 + 59
            if_A(x1,x2,x3) = max{64, x1 - 1, x2 + 53, x3}
            gt_A(x1,x2) = max{68, x1 - 60, x2 - 60}
            not_A(x1) = max{130, x1 - 1}
            id_A(x1) = max{74, x1}
            true_A = 68
            false_A = 63
            zero_A = 0
            plus#!_A(x1,x2) = max{x1 + 56, x2 + 63}
            s!_A(x1) = max{191, x1 - 914}
            if!_A(x1,x2,x3) = max{193, x1 - 852, x2 - 913, x3 - 1334}
            gt!_A(x1,x2) = max{131, x1 - 1399, x2 - 1277}
            not!_A(x1) = max{194, x1 - 851}
            id!_A(x1) = 192
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [1, 2, 3]
      pi(gt) = [1, 2]
      pi(not) = [1]
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: plus#(s(x),y) -> plus#(x,y)
p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: 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),y) -> plus#(x,y)
p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r19, r20, r22, r23, r24

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 - 383, x2 - 201}
            s_A(x1) = max{113, x1 + 32}
            id_A(x1) = max{179, x1 + 154}
            if_A(x1,x2,x3) = max{200, x1 - 1, x2 + 14, x3}
            gt_A(x1,x2) = max{153, x1 - 114, x2 - 38}
            true_A = 153
            false_A = 153
            zero_A = 0
            plus#!_A(x1,x2) = max{202, x2 + 1}
            s!_A(x1) = max{3, x1 - 199}
            id!_A(x1) = max{1, x1 - 583}
            if!_A(x1,x2,x3) = max{2, x1 - 162, x2 - 491, x3 - 457}
            gt!_A(x1,x2) = max{4, x1 - 391, x2 - 114}
            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 - 100, x2 - 59}
            s_A(x1) = max{13, x1}
            id_A(x1) = max{38, x1 + 34}
            if_A(x1,x2,x3) = max{58, x1 - 1, x2, x3}
            gt_A(x1,x2) = max{1, x1 - 14, x2 - 26}
            true_A = 1
            false_A = 1
            zero_A = 0
            plus#!_A(x1,x2) = 0
            s!_A(x1) = max{0, x1 - 40}
            id!_A(x1) = max{2, x1 - 63}
            if!_A(x1,x2,x3) = max{3, x1 - 14, x2 - 39, x3 - 39}
            gt!_A(x1,x2) = max{1, x1 - 39, x2 - 14}
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(id) = [1]
      pi(if) = [1, 2, 3]
      pi(gt) = []
      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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: 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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r19, r20, r22, r23, r24

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 - 126, x2 - 63}
            id_A(x1) = max{60, x1 + 32}
            s_A(x1) = max{59, x1 + 33}
            if_A(x1,x2,x3) = max{60, x1 - 1, x2 + 33, x3}
            gt_A(x1,x2) = max{63, x1 - 60, x2 - 34}
            true_A = 60
            false_A = 0
            zero_A = 0
            plus#!_A(x1,x2) = max{1, x1 - 63, x2 - 453}
            id!_A(x1) = max{3, x1 - 390}
            s!_A(x1) = max{4, x1 - 165}
            if!_A(x1,x2,x3) = max{2, x1 - 95, x2 - 131, x3 - 61}
            gt!_A(x1,x2) = max{5, x1 - 61, x2 - 419}
            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 - 161, x2 - 115}
            id_A(x1) = max{39, x1 + 20}
            s_A(x1) = max{66, x1 + 41}
            if_A(x1,x2,x3) = max{114, x1 - 5, x2 + 26, x3}
            gt_A(x1,x2) = max{73, x1 - 67, x2 - 67}
            true_A = 73
            false_A = 73
            zero_A = 0
            plus#!_A(x1,x2) = max{6, x1 - 40, x2 - 373}
            id!_A(x1) = max{2, x1 - 356}
            s!_A(x1) = max{3, x1 - 304}
            if!_A(x1,x2,x3) = max{5, x1 - 94, x2 - 280, x3 - 308}
            gt!_A(x1,x2) = max{1, x1 - 373, x2 - 260}
            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: app#(cons(x,l),k) -> app#(l,k)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: 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 - 20}
            cons_A(x1,x2) = max{x1 - 77, x2 + 19}
            app#!_A(x1,x2) = max{4, x1 - 2970, x2 - 1}
            cons!_A(x1,x2) = max{3, x1 - 4149, x2 - 2971}
        
        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{x1 - 12, x2 + 5}
            app#!_A(x1,x2) = max{1, x1 - 6, x2 - 1}
            cons!_A(x1,x2) = max{2, x1 - 1367, x2 - 1350}
        
  
    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: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))
r8: app(nil(),k) -> k
r9: app(l,nil()) -> l
r10: app(cons(x,l),k) -> cons(x,app(l,k))
r11: sum(cons(x,nil())) -> cons(x,nil())
r12: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r13: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r14: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r15: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r16: plus(zero(),y) -> y
r17: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r18: id(x) -> x
r19: if(true(),x,y) -> x
r20: if(false(),x,y) -> y
r21: not(x) -> if(x,false(),true())
r22: gt(s(x),zero()) -> true()
r23: gt(zero(),y) -> false()
r24: 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 - 20, x2 - 20}
            s_A(x1) = x1 + 19
            gt#!_A(x1,x2) = max{2, x1 + 1, x2 - 20}
            s!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 24, x2 - 24}
            s_A(x1) = max{42, x1 + 23}
            gt#!_A(x1,x2) = max{107, x1 - 43, x2 - 43}
            s!_A(x1) = x1 + 106
        
  
    partial status:
  
      pi(gt#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.