Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/SK90/2.20)

The rewrite relation of the following TRS is considered.

sum(0) → 0 (1)
sum(s(x)) → +(sqr(s(x)),sum(x)) (2)
sqr(x) → *(x,x) (3)
sum(s(x)) → +(*(s(x),s(x)),sum(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) → c (5)
originates from
sum(0) → 0 (1)
sum#(s(z0)) → c1(sqr#(s(z0)),sum#(z0)) (7)
originates from
sum(s(z0)) → +(sqr(s(z0)),sum(z0)) (6)
sum#(s(z0)) → c2(sum#(z0)) (9)
originates from
sum(s(z0)) → +(*(s(z0),s(z0)),sum(z0)) (8)
sqr#(z0) → c3 (11)
originates from
sqr(z0) → *(z0,z0) (10)
Moreover, we add the following terms to the innermost strategy.
sum#(0)
sum#(s(z0))
sum#(s(z0))
sqr#(z0)

1.1 Usable Rules

We remove the following rules since they are not usable.
sum(0) → 0 (1)
sum(s(z0)) → +(sqr(s(z0)),sum(z0)) (6)
sum(s(z0)) → +(*(s(z0),s(z0)),sum(z0)) (8)
sqr(z0) → *(z0,z0) (10)

1.1.1 Rule Shifting

The rules
sum#(0) → c (5)
sum#(s(z0)) → c2(sum#(z0)) (9)
sqr#(z0) → c3 (11)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c2(x1)] = 1 · x1 + 0
[c3] = 0
[sum#(x1)] = 1 · x1 + 0
[sqr#(x1)] = 1
[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(sqr#(s(z0)),sum#(z0)) (7)
sum#(s(z0)) → c2(sum#(z0)) (9)
sqr#(z0) → c3 (11)

1.1.1.1 Rule Shifting

The rules
sum#(s(z0)) → c1(sqr#(s(z0)),sum#(z0)) (7)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c2(x1)] = 1 · x1 + 0
[c3] = 0
[sum#(x1)] = 1 · x1 + 0
[sqr#(x1)] = 0
[0] = 0
[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(sqr#(s(z0)),sum#(z0)) (7)
sum#(s(z0)) → c2(sum#(z0)) (9)
sqr#(z0) → c3 (11)

1.1.1.1.1 R is empty

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