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, r10, r11, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = max{4, x1}
        app_A(x1,x2) = max{x1 + 21, x2 + 1}
        cons_A(x1,x2) = 12
        sum_A(x1) = max{12, x1 - 7}
        nil_A = 3
        plus_A(x1,x2) = x2 + 24
        s_A(x1) = 0
        if_A(x1,x2,x3) = max{83, x1 - 24, x2 - 48}
        gt_A(x1,x2) = max{24, x1 - 24, x2 - 24}
        not_A(x1) = max{49, x1 - 48}
        id_A(x1) = x1 + 82
        zero_A = 23
  
    base order 2:
  
      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) = x1 + x2 + 6
            sum_A(x1) = x1
            nil_A() = 3
            plus_A(x1,x2) = x2 + 1
            s_A(x1) = 0
            if_A(x1,x2,x3) = 2
            gt_A(x1,x2) = x2 + 1
            not_A(x1) = 1
            id_A(x1) = x1 + 1
            zero_A() = 1
            sum#!_A(x1) = 10
            app!_A(x1,x2) = x1 + x2
            cons!_A(x1,x2) = x2 + 3
            sum!_A(x1) = 8
            nil!_A() = 2
            plus!_A(x1,x2) = 10
            s!_A(x1) = 1
            if!_A(x1,x2,x3) = 11
            gt!_A(x1,x2) = 11
            not!_A(x1) = 1
            id!_A(x1) = x1 + 9
            zero!_A() = 1
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      pi(plus) = []
      pi(s) = []
      pi(if) = []
      pi(gt) = []
      pi(not) = []
      pi(id) = [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: 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:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{x1 + 69, x2 - 1}
        cons_A(x1,x2) = max{x1 + 47, x2 + 68}
  
    base order 2:
  
      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 + 2
            app#!_A(x1,x2) = x1 + 18
            cons!_A(x1,x2) = x2 + 17
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))

and R consists of:

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

The set of usable rules consists of

  r4, r14, r15, r17, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1) = 47
        cons_A(x1,x2) = x2 + 23
        plus_A(x1,x2) = max{12, x2 - 19}
        sum_A(x1) = max{150, x1 - 63}
        nil_A = 125
        if_A(x1,x2,x3) = max{x1 + 154, x2 + 21, x3 + 12}
        true_A = 2
        false_A = 0
        gt_A(x1,x2) = max{12, x1 - 172, x2 - 150}
        s_A(x1) = max{149, x1 + 22}
        zero_A = 149
  
    base order 2:
  
      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 + 1
            plus_A(x1,x2) = x2 + 2
            sum_A(x1) = x1 + 1
            nil_A() = 5
            if_A(x1,x2,x3) = x2 + x3 + 1
            true_A() = 1
            false_A() = 1
            gt_A(x1,x2) = 1
            s_A(x1) = 1
            zero_A() = 1
            sum#!_A(x1) = x1
            cons!_A(x1,x2) = 3
            plus!_A(x1,x2) = 3
            sum!_A(x1) = 5
            nil!_A() = 0
            if!_A(x1,x2,x3) = 4
            true!_A() = 0
            false!_A() = 0
            gt!_A(x1,x2) = 2
            s!_A(x1) = x1 + 6
            zero!_A() = 1
  
    partial status:
  
      pi(sum#) = []
      pi(cons) = []
      pi(plus) = []
      pi(sum) = []
      pi(nil) = []
      pi(if) = [2]
      pi(true) = []
      pi(false) = []
      pi(gt) = []
      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: 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:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{25, x1 + 12, x2 + 9}
        s_A(x1) = x1 + 16
        if_A(x1,x2,x3) = max{15, x2 + 5, x3}
        gt_A(x1,x2) = 0
        not_A(x1) = max{27, x1 - 56}
        id_A(x1) = max{10, x1 + 8}
        true_A = 0
        false_A = 0
        zero_A = 17
  
    base order 2:
  
      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 + 7
            gt_A(x1,x2) = 1
            not_A(x1) = 9
            id_A(x1) = x1 + 7
            true_A() = 1
            false_A() = 1
            zero_A() = 1
            plus#!_A(x1,x2) = 8
            s!_A(x1) = 6
            if!_A(x1,x2,x3) = 0
            gt!_A(x1,x2) = 6
            not!_A(x1) = 7
            id!_A(x1) = 1
            true!_A() = 6
            false!_A() = 0
            zero!_A() = 7
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      pi(if) = [2, 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: 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

  r15

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 - 217}
        s_A(x1) = x1 + 66
        if_A(x1,x2,x3) = max{428, x1 - 347, x2 + 427, x3 + 363}
        gt_A(x1,x2) = max{425, x1 + 135, x2 + 146}
        false_A = 0
  
    base order 2:
  
      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
            if_A(x1,x2,x3) = x3
            gt_A(x1,x2) = x2 + 1
            false_A() = 2
            plus#!_A(x1,x2) = x1 + 3
            id!_A(x1) = 0
            s!_A(x1) = 6
            if!_A(x1,x2,x3) = 5
            gt!_A(x1,x2) = 3
            false!_A() = 0
  
    partial status:
  
      pi(plus#) = []
      pi(id) = []
      pi(s) = []
      pi(if) = []
      pi(gt) = [1, 2]
      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: 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

  r11

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{37, x1 - 37}
        plus_A(x1,x2) = max{74, x1 + 26, x2 + 20}
        zero_A = 155
  
    base order 2:
  
      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) = 0
            s_A(x1) = x1 + 1
            plus_A(x1,x2) = x2 + 2
            zero_A() = 1
            gt#!_A(x1,x2) = x1 + x2 + 6
            s!_A(x1) = x1 + 5
            plus!_A(x1,x2) = 3
            zero!_A() = 7
  
    partial status:
  
      pi(gt#) = []
      pi(s) = []
      pi(plus) = [1]
      pi(zero) = []

The following rules are removed:

  p1

Then no dependency pair remains.