Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/SK90/2.40)

The rewrite relation of the following TRS is considered.

or(true,y)	→	true	(1)
or(x,true)	→	true	(2)
or(false,false)	→	false	(3)
mem(x,nil)	→	false	(4)
mem(x,set(y))	→	=(x,y)	(5)
mem(x,union(y,z))	→	or(mem(x,y),mem(x,z))	(6)

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:

or^#(true,z0)

→

(8)

originates from

or(true,z0)

→

true

(7)

or^#(z0,true)

→

(10)

originates from

or(z0,true)

→

true

(9)

or^#(false,false)

→

(11)

originates from

or(false,false)

→

false

(3)

mem^#(z0,nil)

→

(13)

originates from

mem(z0,nil)

→

false

(12)

mem^#(z0,set(z1))

→

(15)

originates from

mem(z0,set(z1))

→

=(z0,z1)

(14)

mem^#(z0,union(z1,z2))

→

c5(or^#(mem(z0,z1),mem(z0,z2)),mem^#(z0,z1),mem^#(z0,z2))

(17)

originates from

mem(z0,union(z1,z2))

→

or(mem(z0,z1),mem(z0,z2))

(16)

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

or^#(true,z0)

or^#(z0,true)

or^#(false,false)

mem^#(z0,nil)

mem^#(z0,set(z1))

mem^#(z0,union(z1,z2))

1.1 Rule Shifting

The rules

or^#(true,z0)	→	c	(8)
or^#(z0,true)	→	c1	(10)
or^#(false,false)	→	c2	(11)
mem^#(z0,nil)	→	c3	(13)
mem^#(z0,set(z1))	→	c4	(15)

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

[c]	=	0
[c1]	=	0
[c2]	=	0
[c3]	=	0
[c4]	=	0
[c5(x₁, x₂, x₃)]	=	1 · x₁ + 0 + 1 · x₂ + 1 · x₃
[or(x₁, x₂)]	=	1 · x₂ + 0
[mem(x₁, x₂)]	=	1 + 1 · x₁
[or^#(x₁, x₂)]	=	1
[mem^#(x₁, x₂)]	=	1 · x₂ + 0
[nil]	=	1
[false]	=	1
[set(x₁)]	=	1 + 1 · x₁
[=(x₁, x₂)]	=	1 + 1 · x₁
[union(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[true]	=	0

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

or^#(true,z0)	→	c	(8)
or^#(z0,true)	→	c1	(10)
or^#(false,false)	→	c2	(11)
mem^#(z0,nil)	→	c3	(13)
mem^#(z0,set(z1))	→	c4	(15)
mem^#(z0,union(z1,z2))	→	c5(or^#(mem(z0,z1),mem(z0,z2)),mem^#(z0,z1),mem^#(z0,z2))	(17)

1.1.1 Rule Shifting

The rules

mem^#(z0,union(z1,z2))

→

c5(or^#(mem(z0,z1),mem(z0,z2)),mem^#(z0,z1),mem^#(z0,z2))

(17)

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

[c]	=	0
[c1]	=	0
[c2]	=	0
[c3]	=	0
[c4]	=	0
[c5(x₁, x₂, x₃)]	=	1 · x₁ + 0 + 1 · x₂ + 1 · x₃
[or(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[mem(x₁, x₂)]	=	1 + 1 · x₂
[or^#(x₁, x₂)]	=	0
[mem^#(x₁, x₂)]	=	1 · x₂ + 0
[nil]	=	1
[false]	=	1
[set(x₁)]	=	1 + 1 · x₁
[=(x₁, x₂)]	=	1 · x₂ + 0
[union(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[true]	=	1

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

or^#(true,z0)	→	c	(8)
or^#(z0,true)	→	c1	(10)
or^#(false,false)	→	c2	(11)
mem^#(z0,nil)	→	c3	(13)
mem^#(z0,set(z1))	→	c4	(15)
mem^#(z0,union(z1,z2))	→	c5(or^#(mem(z0,z1),mem(z0,z2)),mem^#(z0,z1),mem^#(z0,z2))	(17)

1.1.1.1 R is empty

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