Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/Transformed_CSR_04/Ex6_9_Luc02c_Z)

The rewrite relation of the following TRS is considered.

2nd(cons1(X,cons(Y,Z))) → Y (1)
2nd(cons(X,X1)) → 2nd(cons1(X,activate(X1))) (2)
from(X) → cons(X,n__from(s(X))) (3)
from(X) → n__from(X) (4)
activate(n__from(X)) → from(X) (5)
activate(X) → X (6)
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:
2nd#(cons1(z0,cons(z1,z2))) → c (8)
originates from
2nd(cons1(z0,cons(z1,z2))) → z1 (7)
2nd#(cons(z0,z1)) → c1(2nd#(cons1(z0,activate(z1))),activate#(z1)) (10)
originates from
2nd(cons(z0,z1)) → 2nd(cons1(z0,activate(z1))) (9)
from#(z0) → c2 (12)
originates from
from(z0) → cons(z0,n__from(s(z0))) (11)
from#(z0) → c3 (14)
originates from
from(z0) → n__from(z0) (13)
activate#(n__from(z0)) → c4(from#(z0)) (16)
originates from
activate(n__from(z0)) → from(z0) (15)
activate#(z0) → c5 (18)
originates from
activate(z0) → z0 (17)
Moreover, we add the following terms to the innermost strategy.
2nd#(cons1(z0,cons(z1,z2)))
2nd#(cons(z0,z1))
from#(z0)
from#(z0)
activate#(n__from(z0))
activate#(z0)

1.1 Usable Rules

We remove the following rules since they are not usable.
2nd(cons1(z0,cons(z1,z2))) → z1 (7)
2nd(cons(z0,z1)) → 2nd(cons1(z0,activate(z1))) (9)

1.1.1 Rule Shifting

The rules
2nd#(cons1(z0,cons(z1,z2))) → c (8)
from#(z0) → c2 (12)
from#(z0) → c3 (14)
activate#(n__from(z0)) → c4(from#(z0)) (16)
activate#(z0) → c5 (18)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c2] = 0
[c3] = 0
[c4(x1)] = 1 · x1 + 0
[c5] = 0
[activate(x1)] = 1 + 1 · x1
[from(x1)] = 1 + 1 · x1
[2nd#(x1)] = 1 + 1 · x1
[from#(x1)] = 1 + 1 · x1
[activate#(x1)] = 1 + 1 · x1
[n__from(x1)] = 1 + 1 · x1
[cons(x1, x2)] = 1 + 1 · x1 + 1 · x2
[s(x1)] = 1 + 1 · x1
[cons1(x1, x2)] = 1 · x1 + 0
which has the intended complexity. Here, only the following usable rules have been considered:
2nd#(cons1(z0,cons(z1,z2))) → c (8)
2nd#(cons(z0,z1)) → c1(2nd#(cons1(z0,activate(z1))),activate#(z1)) (10)
from#(z0) → c2 (12)
from#(z0) → c3 (14)
activate#(n__from(z0)) → c4(from#(z0)) (16)
activate#(z0) → c5 (18)

1.1.1.1 Rule Shifting

The rules
2nd#(cons(z0,z1)) → c1(2nd#(cons1(z0,activate(z1))),activate#(z1)) (10)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c2] = 0
[c3] = 0
[c4(x1)] = 1 · x1 + 0
[c5] = 0
[activate(x1)] = 3
[from(x1)] = 3 + 3 · x1
[2nd#(x1)] = 1 · x1 + 0
[from#(x1)] = 0
[activate#(x1)] = 0
[n__from(x1)] = 3 + 1 · x1
[cons(x1, x2)] = 1 + 1 · x1
[s(x1)] = 3 + 1 · x1
[cons1(x1, x2)] = 0
which has the intended complexity. Here, only the following usable rules have been considered:
2nd#(cons1(z0,cons(z1,z2))) → c (8)
2nd#(cons(z0,z1)) → c1(2nd#(cons1(z0,activate(z1))),activate#(z1)) (10)
from#(z0) → c2 (12)
from#(z0) → c3 (14)
activate#(n__from(z0)) → c4(from#(z0)) (16)
activate#(z0) → c5 (18)

1.1.1.1.1 R is empty

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