Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/AG01/#3.2)

The rewrite relation of the following TRS is considered.

pred(s(x))	→	x	(1)
minus(x,0)	→	x	(2)
minus(x,s(y))	→	pred(minus(x,y))	(3)
quot(0,s(y))	→	0	(4)
quot(s(x),s(y))	→	s(quot(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:

pred^#(s(z0))

→

(7)

originates from

pred(s(z0))

→

(6)

minus^#(z0,0)

→

(9)

originates from

minus(z0,0)

→

(8)

minus^#(z0,s(z1))

→

c2(pred^#(minus(z0,z1)),minus^#(z0,z1))

(11)

originates from

minus(z0,s(z1))

→

pred(minus(z0,z1))

(10)

quot^#(0,s(z0))

→

(13)

originates from

quot(0,s(z0))

→

(12)

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

→

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

(15)

originates from

quot(s(z0),s(z1))

→

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

(14)

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

pred^#(s(z0))

minus^#(z0,0)

minus^#(z0,s(z1))

quot^#(0,s(z0))

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

1.1 Usable Rules

We remove the following rules since they are not usable.

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

1.1.1 Rule Shifting

The rules

quot^#(0,s(z0))

→

(13)

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

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

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

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

1.1.1.1 Rule Shifting

The rules

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

→

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

(15)

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

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

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

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

1.1.1.1.1 Rule Shifting

The rules

minus^#(z0,0)

→

(9)

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

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

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

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

1.1.1.1.1.1 Rule Shifting

The rules

minus^#(z0,s(z1))

→

c2(pred^#(minus(z0,z1)),minus^#(z0,z1))

(11)

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

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

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

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

1.1.1.1.1.1.1 Rule Shifting

The rules

pred^#(s(z0))

→

(7)

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

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

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

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

1.1.1.1.1.1.1.1 R is empty

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