Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/Der95/08)

The rewrite relation of the following TRS is considered.

D(t)	→	1	(1)
D(constant)	→	0	(2)
D(+(x,y))	→	+(D(x),D(y))	(3)
D(*(x,y))	→	+((y,D(x)),(x,D(y)))	(4)
D(-(x,y))	→	-(D(x),D(y))	(5)

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:

D^#(t)

→

(6)

originates from

D(t)

→

(1)

D^#(constant)

→

(7)

originates from

D(constant)

→

(2)

D^#(+(z0,z1))

→

c2(D^#(z0),D^#(z1))

(9)

originates from

D(+(z0,z1))

→

+(D(z0),D(z1))

(8)

D^#(*(z0,z1))

→

c3(D^#(z0),D^#(z1))

(11)

originates from

D(*(z0,z1))

→

+(*(z1,D(z0)),*(z0,D(z1)))

(10)

D^#(-(z0,z1))

→

c4(D^#(z0),D^#(z1))

(13)

originates from

D(-(z0,z1))

→

-(D(z0),D(z1))

(12)

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

D^#(t)

D^#(constant)

D^#(+(z0,z1))

D^#(*(z0,z1))

D^#(-(z0,z1))

1.1 Usable Rules

We remove the following rules since they are not usable.

D(t)	→	1	(1)
D(constant)	→	0	(2)
D(+(z0,z1))	→	+(D(z0),D(z1))	(8)
D(*(z0,z1))	→	+((z1,D(z0)),(z0,D(z1)))	(10)
D(-(z0,z1))	→	-(D(z0),D(z1))	(12)

1.1.1 Rule Shifting

The rules

D^#(+(z0,z1))	→	c2(D^#(z0),D^#(z1))	(9)
D^#(*(z0,z1))	→	c3(D^#(z0),D^#(z1))	(11)
D^#(-(z0,z1))	→	c4(D^#(z0),D^#(z1))	(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(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c4(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[D^#(x₁)]	=	1 · x₁ + 0
[t]	=	0
[constant]	=	0
[+(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[*(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂
[-(x₁, x₂)]	=	1 + 1 · x₁ + 1 · x₂

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

D^#(t)	→	c	(6)
D^#(constant)	→	c1	(7)
D^#(+(z0,z1))	→	c2(D^#(z0),D^#(z1))	(9)
D^#(*(z0,z1))	→	c3(D^#(z0),D^#(z1))	(11)
D^#(-(z0,z1))	→	c4(D^#(z0),D^#(z1))	(13)

1.1.1.1 Rule Shifting

The rules

D^#(t)	→	c	(6)
D^#(constant)	→	c1	(7)

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(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[c4(x₁, x₂)]	=	1 · x₁ + 0 + 1 · x₂
[D^#(x₁)]	=	3 + 3 · x₁
[t]	=	1
[constant]	=	1
[+(x₁, x₂)]	=	2 + 1 · x₁ + 1 · x₂
[*(x₁, x₂)]	=	3 + 1 · x₁ + 1 · x₂
[-(x₁, x₂)]	=	2 + 1 · x₁ + 1 · x₂

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

D^#(t)	→	c	(6)
D^#(constant)	→	c1	(7)
D^#(+(z0,z1))	→	c2(D^#(z0),D^#(z1))	(9)
D^#(*(z0,z1))	→	c3(D^#(z0),D^#(z1))	(11)
D^#(-(z0,z1))	→	c4(D^#(z0),D^#(z1))	(13)

1.1.1.1.1 R is empty

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