Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/SK90/2.37)

The rewrite relation of the following TRS is considered.

and(not(not(x)),y,not(z))

→

and(y,band(x,z),x)

(1)

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:

and^#(not(not(z0)),z1,not(z2))

→

c(and^#(z1,band(z0,z2),z0))

(3)

originates from

and(not(not(z0)),z1,not(z2))

→

and(z1,band(z0,z2),z0)

(2)

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

and^#(not(not(z0)),z1,not(z2))

1.1 Usable Rules

We remove the following rules since they are not usable.

and(not(not(z0)),z1,not(z2))

→

and(z1,band(z0,z2),z0)

(2)

1.1.1 Rule Shifting

The rules

and^#(not(not(z0)),z1,not(z2))

→

c(and^#(z1,band(z0,z2),z0))

(3)

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

[c(x₁)]	=	1 · x₁ + 0
[and^#(x₁, x₂, x₃)]	=	1 · x₁ + 0 + 3 · x₂
[not(x₁)]	=	1
[band(x₁, x₂)]	=	0

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

and^#(not(not(z0)),z1,not(z2))

→

c(and^#(z1,band(z0,z2),z0))

(3)

1.1.1.1 R is empty

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