Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/AG01/#3.1)

The rewrite relation of the following TRS is considered.

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

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)

→

(6)

originates from

minus(z0,0)

→

(5)

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

→

c1(minus^#(z0,z1))

(8)

originates from

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

→

minus(z0,z1)

(7)

quot^#(0,s(z0))

→

(10)

originates from

quot(0,s(z0))

→

(9)

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

→

c3(quot^#(minus(z0,z1),s(z1)),minus^#(z0,z1))

(12)

originates from

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

→

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

(11)

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

minus^#(z0,0)

minus^#(s(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	(9)
quot(s(z0),s(z1))	→	s(quot(minus(z0,z1),s(z1)))	(11)

1.1.1 Rule Shifting

The rules

minus^#(z0,0)	→	c	(6)
quot^#(0,s(z0))	→	c2	(10)

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

[c]	=	0
[c1(x₁)]	=	1 · x₁ + 0
[c2]	=	0
[c3(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[minus(x₁, x₂)]	=	1 · x₁ + 0
[minus^#(x₁, x₂)]	=	1
[quot^#(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[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	(6)
minus^#(s(z0),s(z1))	→	c1(minus^#(z0,z1))	(8)
quot^#(0,s(z0))	→	c2	(10)
quot^#(s(z0),s(z1))	→	c3(quot^#(minus(z0,z1),s(z1)),minus^#(z0,z1))	(12)
minus(s(z0),s(z1))	→	minus(z0,z1)	(7)
minus(z0,0)	→	z0	(5)

1.1.1.1 Rule Shifting

The rules

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

→

c3(quot^#(minus(z0,z1),s(z1)),minus^#(z0,z1))

(12)

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

[c]	=	0
[c1(x₁)]	=	1 · x₁ + 0
[c2]	=	0
[c3(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[minus(x₁, x₂)]	=	1 · 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:

minus^#(z0,0)	→	c	(6)
minus^#(s(z0),s(z1))	→	c1(minus^#(z0,z1))	(8)
quot^#(0,s(z0))	→	c2	(10)
quot^#(s(z0),s(z1))	→	c3(quot^#(minus(z0,z1),s(z1)),minus^#(z0,z1))	(12)
minus(s(z0),s(z1))	→	minus(z0,z1)	(7)
minus(z0,0)	→	z0	(5)

1.1.1.1.1 Rule Shifting

The rules

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

→

c1(minus^#(z0,z1))

(8)

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

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

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

minus^#(z0,0)	→	c	(6)
minus^#(s(z0),s(z1))	→	c1(minus^#(z0,z1))	(8)
quot^#(0,s(z0))	→	c2	(10)
quot^#(s(z0),s(z1))	→	c3(quot^#(minus(z0,z1),s(z1)),minus^#(z0,z1))	(12)
minus(s(z0),s(z1))	→	minus(z0,z1)	(7)
minus(z0,0)	→	z0	(5)

1.1.1.1.1.1 R is empty

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