Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/Rubio_04/gm)

The rewrite relation of the following TRS is considered.

minus(X,0)	→	X	(1)
minus(s(X),s(Y))	→	p(minus(X,Y))	(2)
p(s(X))	→	X	(3)
div(0,s(Y))	→	0	(4)
div(s(X),s(Y))	→	s(div(minus(X,Y),s(Y)))	(5)

The evaluation strategy is innermost.

Property / Task

Determine bounds on the runtime complexity.

Answer / Result

An upperbound for the complexity is O(n²).

Proof (by AProVE @ termCOMP 2023)

1 Dependency Tuples

We get the following set of dependency tuples:

minus^#(z0,0)

→

(7)

originates from

minus(z0,0)

→

(6)

minus^#(s(z0),s(z1))

→

c1(p^#(minus(z0,z1)),minus^#(z0,z1))

(9)

originates from

minus(s(z0),s(z1))

→

p(minus(z0,z1))

(8)

p^#(s(z0))

→

(11)

originates from

p(s(z0))

→

(10)

div^#(0,s(z0))

→

(13)

originates from

div(0,s(z0))

→

(12)

div^#(s(z0),s(z1))

→

c4(div^#(minus(z0,z1),s(z1)),minus^#(z0,z1))

(15)

originates from

div(s(z0),s(z1))

→

s(div(minus(z0,z1),s(z1)))

(14)

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

minus^#(z0,0)

minus^#(s(z0),s(z1))

p^#(s(z0))

div^#(0,s(z0))

div^#(s(z0),s(z1))

1.1 Usable Rules

We remove the following rules since they are not usable.

div(0,s(z0))	→	0	(12)
div(s(z0),s(z1))	→	s(div(minus(z0,z1),s(z1)))	(14)

1.1.1 Rule Shifting

The rules

div^#(0,s(z0))

→

(13)

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

[c]	=	0
[c1(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c2]	=	0
[c3]	=	0
[c4(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[minus(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[p(x₁)]	=	1 + 1 · x₁
[minus^#(x₁, x₂)]	=	0
[p^#(x₁)]	=	0
[div^#(x₁, x₂)]	=	1 · x₂ + 0
[0]	=	1
[s(x₁)]	=	1 + 1 · x₁

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

minus^#(z0,0)	→	c	(7)
minus^#(s(z0),s(z1))	→	c1(p^#(minus(z0,z1)),minus^#(z0,z1))	(9)
p^#(s(z0))	→	c2	(11)
div^#(0,s(z0))	→	c3	(13)
div^#(s(z0),s(z1))	→	c4(div^#(minus(z0,z1),s(z1)),minus^#(z0,z1))	(15)

1.1.1.1 Rule Shifting

The rules

minus^#(z0,0)	→	c	(7)
div^#(s(z0),s(z1))	→	c4(div^#(minus(z0,z1),s(z1)),minus^#(z0,z1))	(15)

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

[c]	=	0
[c1(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c2]	=	0
[c3]	=	0
[c4(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[minus(x₁, x₂)]	=	1 · x₁ + 0
[p(x₁)]	=	1 · x₁ + 0
[minus^#(x₁, x₂)]	=	1
[p^#(x₁)]	=	0
[div^#(x₁, x₂)]	=	1 · x₁ + 0
[0]	=	0
[s(x₁)]	=	2 + 1 · x₁

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

minus^#(z0,0)	→	c	(7)
minus^#(s(z0),s(z1))	→	c1(p^#(minus(z0,z1)),minus^#(z0,z1))	(9)
p^#(s(z0))	→	c2	(11)
div^#(0,s(z0))	→	c3	(13)
div^#(s(z0),s(z1))	→	c4(div^#(minus(z0,z1),s(z1)),minus^#(z0,z1))	(15)
p(s(z0))	→	z0	(10)
minus(z0,0)	→	z0	(6)
minus(s(z0),s(z1))	→	p(minus(z0,z1))	(8)

1.1.1.1.1 Rule Shifting

The rules

minus^#(s(z0),s(z1))	→	c1(p^#(minus(z0,z1)),minus^#(z0,z1))	(9)
p^#(s(z0))	→	c2	(11)

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

[c]	=	0
[c1(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c2]	=	0
[c3]	=	0
[c4(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[minus(x₁, x₂)]	=	1 · x₁ + 0
[p(x₁)]	=	1 · x₁ + 0
[minus^#(x₁, x₂)]	=	1 + 2 · x₁
[p^#(x₁)]	=	1
[div^#(x₁, x₂)]	=	1 · x₁ · x₁ + 0
[0]	=	2
[s(x₁)]	=	1 + 1 · x₁

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

minus^#(z0,0)	→	c	(7)
minus^#(s(z0),s(z1))	→	c1(p^#(minus(z0,z1)),minus^#(z0,z1))	(9)
p^#(s(z0))	→	c2	(11)
div^#(0,s(z0))	→	c3	(13)
div^#(s(z0),s(z1))	→	c4(div^#(minus(z0,z1),s(z1)),minus^#(z0,z1))	(15)
p(s(z0))	→	z0	(10)
minus(z0,0)	→	z0	(6)
minus(s(z0),s(z1))	→	p(minus(z0,z1))	(8)

1.1.1.1.1.1 R is empty

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