Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/Frederiksen_Glenstrup/member)

The relative rewrite relation R/S is considered where R is the following TRS

member(x',Cons(x,xs))	→	member[Ite][True][Ite](!EQ(x',x),x',Cons(x,xs))	(1)
member(x,Nil)	→	False	(2)
notEmpty(Cons(x,xs))	→	True	(3)
notEmpty(Nil)	→	False	(4)
goal(x,xs)	→	member(x,xs)	(5)

and S is the following TRS.

!EQ(S(x),S(y))	→	!EQ(x,y)	(6)
!EQ(0,S(y))	→	False	(7)
!EQ(S(x),0)	→	False	(8)
!EQ(0,0)	→	True	(9)
member[Ite][True][Ite](False,x',Cons(x,xs))	→	member(x',xs)	(10)
member[Ite][True][Ite](True,x,xs)	→	True	(11)

The evaluation strategy is innermost.

Property / Task

Determine bounds on the runtime complexity.

Answer / Result

An upperbound for the complexity is O(n).

Proof (by AProVE @ termCOMP 2023)

1 Dependency Tuples

We get the following set of dependency tuples:

member^#(z0,Cons(z1,z2))

→

c6(member[Ite][True][Ite]^#(!EQ(z0,z1),z0,Cons(z1,z2)),!EQ^#(z0,z1))

(13)

originates from

member(z0,Cons(z1,z2))

→

member[Ite][True][Ite](!EQ(z0,z1),z0,Cons(z1,z2))

(12)

member^#(z0,Nil)

→

(15)

originates from

member(z0,Nil)

→

False

(14)

notEmpty^#(Cons(z0,z1))

→

(17)

originates from

notEmpty(Cons(z0,z1))

→

True

(16)

notEmpty^#(Nil)

→

(18)

originates from

notEmpty(Nil)

→

False

(4)

goal^#(z0,z1)

→

c10(member^#(z0,z1))

(20)

originates from

goal(z0,z1)

→

member(z0,z1)

(19)

!EQ^#(S(z0),S(z1))

→

c(!EQ^#(z0,z1))

(22)

originates from

!EQ(S(z0),S(z1))

→

!EQ(z0,z1)

(21)

!EQ^#(0,S(z0))

→

(24)

originates from

!EQ(0,S(z0))

→

False

(23)

!EQ^#(S(z0),0)

→

(26)

originates from

!EQ(S(z0),0)

→

False

(25)

!EQ^#(0,0)

→

(27)

originates from

!EQ(0,0)

→

True

(9)

member[Ite][True][Ite]^#(False,z0,Cons(z1,z2))

→

c4(member^#(z0,z2))

(29)

originates from

member[Ite][True][Ite](False,z0,Cons(z1,z2))

→

member(z0,z2)

(28)

member[Ite][True][Ite]^#(True,z0,z1)

→

(31)

originates from

member[Ite][True][Ite](True,z0,z1)

→

True

(30)

Moreover, we add the following terms to the innermost strategy.

!EQ^#(S(z0),S(z1))

!EQ^#(0,S(z0))

!EQ^#(S(z0),0)

!EQ^#(0,0)

member[Ite][True][Ite]^#(False,z0,Cons(z1,z2))

member[Ite][True][Ite]^#(True,z0,z1)

member^#(z0,Cons(z1,z2))

member^#(z0,Nil)

notEmpty^#(Cons(z0,z1))

notEmpty^#(Nil)

goal^#(z0,z1)

1.1 Usable Rules

We remove the following rules since they are not usable.

member[Ite][True][Ite](False,z0,Cons(z1,z2))	→	member(z0,z2)	(28)
member[Ite][True][Ite](True,z0,z1)	→	True	(30)
member(z0,Cons(z1,z2))	→	member[Ite][True][Ite](!EQ(z0,z1),z0,Cons(z1,z2))	(12)
member(z0,Nil)	→	False	(14)
notEmpty(Cons(z0,z1))	→	True	(16)
notEmpty(Nil)	→	False	(4)
goal(z0,z1)	→	member(z0,z1)	(19)

1.1.1 Rule Shifting

The rules

member^#(z0,Nil)	→	c7	(15)
notEmpty^#(Cons(z0,z1))	→	c8	(17)
notEmpty^#(Nil)	→	c9	(18)

are strictly oriented by the following linear polynomial interpretation over the naturals

[c(x₁)]	=	1 · x₁ + 0
[c1]	=	0
[c2]	=	0
[c3]	=	0
[c4(x₁)]	=	1 · x₁ + 0
[c5]	=	0
[c6(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c7]	=	0
[c8]	=	0
[c9]	=	0
[c10(x₁)]	=	1 · x₁ + 0
[!EQ(x₁, x₂)]	=	0
[!EQ^#(x₁, x₂)]	=	0
[member[Ite][True][Ite]^#(x₁, x₂, x₃)]	=	1 · x₁ + 0 + 1 · x₂ + 1 · x₃
[member^#(x₁, x₂)]	=	1 + 1 · x₁
[notEmpty^#(x₁)]	=	1 · x₁ + 0
[goal^#(x₁, x₂)]	=	1 + 1 · x₁
[S(x₁)]	=	1 + 1 · x₁
[0]	=	1
[False]	=	0
[True]	=	0
[Cons(x₁, x₂)]	=	1
[Nil]	=	1

which has the intended complexity. Here, only the following usable rules have been considered:

!EQ^#(S(z0),S(z1))	→	c(!EQ^#(z0,z1))	(22)
!EQ^#(0,S(z0))	→	c1	(24)
!EQ^#(S(z0),0)	→	c2	(26)
!EQ^#(0,0)	→	c3	(27)
member[Ite][True][Ite]^#(False,z0,Cons(z1,z2))	→	c4(member^#(z0,z2))	(29)
member[Ite][True][Ite]^#(True,z0,z1)	→	c5	(31)
member^#(z0,Cons(z1,z2))	→	c6(member[Ite][True][Ite]^#(!EQ(z0,z1),z0,Cons(z1,z2)),!EQ^#(z0,z1))	(13)
member^#(z0,Nil)	→	c7	(15)
notEmpty^#(Cons(z0,z1))	→	c8	(17)
notEmpty^#(Nil)	→	c9	(18)
goal^#(z0,z1)	→	c10(member^#(z0,z1))	(20)
!EQ(0,S(z0))	→	False	(23)
!EQ(S(z0),0)	→	False	(25)
!EQ(S(z0),S(z1))	→	!EQ(z0,z1)	(21)
!EQ(0,0)	→	True	(9)

1.1.1.1 Rule Shifting

The rules

goal^#(z0,z1)

→

c10(member^#(z0,z1))

(20)

are strictly oriented by the following linear polynomial interpretation over the naturals

[c(x₁)]	=	1 · x₁ + 0
[c1]	=	0
[c2]	=	0
[c3]	=	0
[c4(x₁)]	=	1 · x₁ + 0
[c5]	=	0
[c6(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c7]	=	0
[c8]	=	0
[c9]	=	0
[c10(x₁)]	=	1 · x₁ + 0
[!EQ(x₁, x₂)]	=	3
[!EQ^#(x₁, x₂)]	=	0
[member[Ite][True][Ite]^#(x₁, x₂, x₃)]	=	3 · x₂ + 0
[member^#(x₁, x₂)]	=	3 · x₁ + 0
[notEmpty^#(x₁)]	=	0
[goal^#(x₁, x₂)]	=	1 + 3 · x₁
[S(x₁)]	=	3 + 1 · x₁
[0]	=	3
[False]	=	3
[True]	=	3
[Cons(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[Nil]	=	0

which has the intended complexity. Here, only the following usable rules have been considered:

!EQ^#(S(z0),S(z1))	→	c(!EQ^#(z0,z1))	(22)
!EQ^#(0,S(z0))	→	c1	(24)
!EQ^#(S(z0),0)	→	c2	(26)
!EQ^#(0,0)	→	c3	(27)
member[Ite][True][Ite]^#(False,z0,Cons(z1,z2))	→	c4(member^#(z0,z2))	(29)
member[Ite][True][Ite]^#(True,z0,z1)	→	c5	(31)
member^#(z0,Cons(z1,z2))	→	c6(member[Ite][True][Ite]^#(!EQ(z0,z1),z0,Cons(z1,z2)),!EQ^#(z0,z1))	(13)
member^#(z0,Nil)	→	c7	(15)
notEmpty^#(Cons(z0,z1))	→	c8	(17)
notEmpty^#(Nil)	→	c9	(18)
goal^#(z0,z1)	→	c10(member^#(z0,z1))	(20)

1.1.1.1.1 Rule Shifting

The rules

member^#(z0,Cons(z1,z2))

→

c6(member[Ite][True][Ite]^#(!EQ(z0,z1),z0,Cons(z1,z2)),!EQ^#(z0,z1))

(13)

are strictly oriented by the following linear polynomial interpretation over the naturals

[c(x₁)]	=	1 · x₁ + 0
[c1]	=	0
[c2]	=	0
[c3]	=	0
[c4(x₁)]	=	1 · x₁ + 0
[c5]	=	0
[c6(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c7]	=	0
[c8]	=	0
[c9]	=	0
[c10(x₁)]	=	1 · x₁ + 0
[!EQ(x₁, x₂)]	=	3
[!EQ^#(x₁, x₂)]	=	0
[member[Ite][True][Ite]^#(x₁, x₂, x₃)]	=	3 · x₂ + 0 + 2 · x₃
[member^#(x₁, x₂)]	=	1 + 3 · x₁ + 2 · x₂
[notEmpty^#(x₁)]	=	0
[goal^#(x₁, x₂)]	=	1 + 3 · x₁ + 2 · x₂
[S(x₁)]	=	3 + 1 · x₁
[0]	=	3
[False]	=	3
[True]	=	3
[Cons(x₁, x₂)]	=	1 + 1 · x₂
[Nil]	=	0

which has the intended complexity. Here, only the following usable rules have been considered:

!EQ^#(S(z0),S(z1))	→	c(!EQ^#(z0,z1))	(22)
!EQ^#(0,S(z0))	→	c1	(24)
!EQ^#(S(z0),0)	→	c2	(26)
!EQ^#(0,0)	→	c3	(27)
member[Ite][True][Ite]^#(False,z0,Cons(z1,z2))	→	c4(member^#(z0,z2))	(29)
member[Ite][True][Ite]^#(True,z0,z1)	→	c5	(31)
member^#(z0,Cons(z1,z2))	→	c6(member[Ite][True][Ite]^#(!EQ(z0,z1),z0,Cons(z1,z2)),!EQ^#(z0,z1))	(13)
member^#(z0,Nil)	→	c7	(15)
notEmpty^#(Cons(z0,z1))	→	c8	(17)
notEmpty^#(Nil)	→	c9	(18)
goal^#(z0,z1)	→	c10(member^#(z0,z1))	(20)

1.1.1.1.1.1 R is empty

There are no rules in the TRS R. Hence, R/S has complexity O(1).