Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/SK90/2.17)

The rewrite relation of the following TRS is considered.

sum(0)	→	0	(1)
sum(s(x))	→	+(sum(x),s(x))	(2)
sum1(0)	→	0	(3)
sum1(s(x))	→	s(+(sum1(x),+(x,x)))	(4)

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:

sum^#(0)

→

(5)

originates from

sum(0)

→

(1)

sum^#(s(z0))

→

c1(sum^#(z0))

(7)

originates from

sum(s(z0))

→

+(sum(z0),s(z0))

(6)

sum1^#(0)

→

(8)

originates from

sum1(0)

→

(3)

sum1^#(s(z0))

→

c3(sum1^#(z0))

(10)

originates from

sum1(s(z0))

→

s(+(sum1(z0),+(z0,z0)))

(9)

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

sum^#(0)

sum^#(s(z0))

sum1^#(0)

sum1^#(s(z0))

1.1 Usable Rules

We remove the following rules since they are not usable.

sum(0)	→	0	(1)
sum(s(z0))	→	+(sum(z0),s(z0))	(6)
sum1(0)	→	0	(3)
sum1(s(z0))	→	s(+(sum1(z0),+(z0,z0)))	(9)

1.1.1 Rule Shifting

The rules

sum^#(0)	→	c	(5)
sum^#(s(z0))	→	c1(sum^#(z0))	(7)
sum1^#(0)	→	c2	(8)
sum1^#(s(z0))	→	c3(sum1^#(z0))	(10)

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

[c]	=	0
[c1(x₁)]	=	1 · x₁ + 0
[c2]	=	0
[c3(x₁)]	=	1 · x₁ + 0
[sum^#(x₁)]	=	1 · x₁ + 0
[sum1^#(x₁)]	=	1 · x₁ + 0
[0]	=	1
[s(x₁)]	=	1 + 1 · x₁

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

sum^#(0)	→	c	(5)
sum^#(s(z0))	→	c1(sum^#(z0))	(7)
sum1^#(0)	→	c2	(8)
sum1^#(s(z0))	→	c3(sum1^#(z0))	(10)

1.1.1.1 R is empty

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