Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/Rubio_04/prov)

The rewrite relation of the following TRS is considered.

ackin(s(X),s(Y)) → u21(ackin(s(X),Y),X) (1)
u21(ackout(X),Y) → u22(ackin(Y,X)) (2)
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:
ackin#(s(z0),s(z1)) → c(u21#(ackin(s(z0),z1),z0),ackin#(s(z0),z1)) (4)
originates from
ackin(s(z0),s(z1)) → u21(ackin(s(z0),z1),z0) (3)
u21#(ackout(z0),z1) → c1(ackin#(z1,z0)) (6)
originates from
u21(ackout(z0),z1) → u22(ackin(z1,z0)) (5)
Moreover, we add the following terms to the innermost strategy.
ackin#(s(z0),s(z1))
u21#(ackout(z0),z1)

1.1 Rule Shifting

The rules
u21#(ackout(z0),z1) → c1(ackin#(z1,z0)) (6)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c1(x1)] = 1 · x1 + 0
[ackin(x1, x2)] = 0
[u21(x1, x2)] = 2 · x1 + 0
[ackin#(x1, x2)] = 0
[u21#(x1, x2)] = 1 · x1 + 0
[s(x1)] = 0
[ackout(x1)] = 3 + 1 · x1
[u22(x1)] = 0
which has the intended complexity. Here, only the following usable rules have been considered:
ackin#(s(z0),s(z1)) → c(u21#(ackin(s(z0),z1),z0),ackin#(s(z0),z1)) (4)
u21#(ackout(z0),z1) → c1(ackin#(z1,z0)) (6)
u21(ackout(z0),z1) → u22(ackin(z1,z0)) (5)
ackin(s(z0),s(z1)) → u21(ackin(s(z0),z1),z0) (3)

1.1.1 Rule Shifting

The rules
ackin#(s(z0),s(z1)) → c(u21#(ackin(s(z0),z1),z0),ackin#(s(z0),z1)) (4)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c1(x1)] = 1 · x1 + 0
[ackin(x1, x2)] = 0
[u21(x1, x2)] = 2 · x1 + 0
[ackin#(x1, x2)] = 3 + 1 · x2
[u21#(x1, x2)] = 1 · x1 + 0
[s(x1)] = 1 + 1 · x1
[ackout(x1)] = 3 + 1 · x1
[u22(x1)] = 0
which has the intended complexity. Here, only the following usable rules have been considered:
ackin#(s(z0),s(z1)) → c(u21#(ackin(s(z0),z1),z0),ackin#(s(z0),z1)) (4)
u21#(ackout(z0),z1) → c1(ackin#(z1,z0)) (6)
u21(ackout(z0),z1) → u22(ackin(z1,z0)) (5)
ackin(s(z0),s(z1)) → u21(ackin(s(z0),z1),z0) (3)

1.1.1.1 R is empty

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