Certification Problem

Input (TPDB Runtime_Complexity_Innermost_Rewriting/SK90/2.29)

The rewrite relation of the following TRS is considered.

prime(0) → false (1)
prime(s(0)) → false (2)
prime(s(s(x))) → prime1(s(s(x)),s(x)) (3)
prime1(x,0) → false (4)
prime1(x,s(0)) → true (5)
prime1(x,s(s(y))) → and(not(divp(s(s(y)),x)),prime1(x,s(y))) (6)
divp(x,y) → =(rem(x,y),0) (7)
The evaluation strategy is innermost.

Property / Task

Determine bounds on the runtime complexity.

Answer / Result

An upperbound for the complexity is O(n2).

Proof (by AProVE @ termCOMP 2023)

1 Dependency Tuples

We get the following set of dependency tuples:
prime#(0) → c (8)
originates from
prime(0) → false (1)
prime#(s(0)) → c1 (9)
originates from
prime(s(0)) → false (2)
prime#(s(s(z0))) → c2(prime1#(s(s(z0)),s(z0))) (11)
originates from
prime(s(s(z0))) → prime1(s(s(z0)),s(z0)) (10)
prime1#(z0,0) → c3 (13)
originates from
prime1(z0,0) → false (12)
prime1#(z0,s(0)) → c4 (15)
originates from
prime1(z0,s(0)) → true (14)
prime1#(z0,s(s(z1))) → c5(divp#(s(s(z1)),z0),prime1#(z0,s(z1))) (17)
originates from
prime1(z0,s(s(z1))) → and(not(divp(s(s(z1)),z0)),prime1(z0,s(z1))) (16)
divp#(z0,z1) → c6 (19)
originates from
divp(z0,z1) → =(rem(z0,z1),0) (18)
Moreover, we add the following terms to the innermost strategy.
prime#(0)
prime#(s(0))
prime#(s(s(z0)))
prime1#(z0,0)
prime1#(z0,s(0))
prime1#(z0,s(s(z1)))
divp#(z0,z1)

1.1 Usable Rules

We remove the following rules since they are not usable.
prime(0) → false (1)
prime(s(0)) → false (2)
prime(s(s(z0))) → prime1(s(s(z0)),s(z0)) (10)
prime1(z0,0) → false (12)
prime1(z0,s(0)) → true (14)
prime1(z0,s(s(z1))) → and(not(divp(s(s(z1)),z0)),prime1(z0,s(z1))) (16)
divp(z0,z1) → =(rem(z0,z1),0) (18)

1.1.1 Rule Shifting

The rules
prime#(0) → c (8)
prime#(s(0)) → c1 (9)
prime#(s(s(z0))) → c2(prime1#(s(s(z0)),s(z0))) (11)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1] = 0
[c2(x1)] = 1 · x1 + 0
[c3] = 0
[c4] = 0
[c5(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c6] = 0
[prime#(x1)] = 1
[prime1#(x1, x2)] = 0
[divp#(x1, x2)] = 0
[0] = 0
[s(x1)] = 0
which has the intended complexity. Here, only the following usable rules have been considered:
prime#(0) → c (8)
prime#(s(0)) → c1 (9)
prime#(s(s(z0))) → c2(prime1#(s(s(z0)),s(z0))) (11)
prime1#(z0,0) → c3 (13)
prime1#(z0,s(0)) → c4 (15)
prime1#(z0,s(s(z1))) → c5(divp#(s(s(z1)),z0),prime1#(z0,s(z1))) (17)
divp#(z0,z1) → c6 (19)

1.1.1.1 Rule Shifting

The rules
prime1#(z0,s(0)) → c4 (15)
prime1#(z0,s(s(z1))) → c5(divp#(s(s(z1)),z0),prime1#(z0,s(z1))) (17)
are strictly oriented by the following non-linear polynomial interpretation over the naturals
[c] = 0
[c1] = 0
[c2(x1)] = 1 · x1 + 0
[c3] = 0
[c4] = 0
[c5(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c6] = 0
[prime#(x1)] = 2 · x1 · x1 + 0
[prime1#(x1, x2)] = 1 · x1 + 0 + 1 · x2 + 2 · x2 · x2
[divp#(x1, x2)] = 1 · x1 + 0
[0] = 0
[s(x1)] = 2 + 1 · x1
which has the intended complexity. Here, only the following usable rules have been considered:
prime#(0) → c (8)
prime#(s(0)) → c1 (9)
prime#(s(s(z0))) → c2(prime1#(s(s(z0)),s(z0))) (11)
prime1#(z0,0) → c3 (13)
prime1#(z0,s(0)) → c4 (15)
prime1#(z0,s(s(z1))) → c5(divp#(s(s(z1)),z0),prime1#(z0,s(z1))) (17)
divp#(z0,z1) → c6 (19)

1.1.1.1.1 Rule Shifting

The rules
prime1#(z0,0) → c3 (13)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1] = 0
[c2(x1)] = 1 · x1 + 0
[c3] = 0
[c4] = 0
[c5(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c6] = 0
[prime#(x1)] = 1 + 1 · x1
[prime1#(x1, x2)] = 1 · x1 + 0 + 1 · x2
[divp#(x1, x2)] = 0
[0] = 1
[s(x1)] = 1
which has the intended complexity. Here, only the following usable rules have been considered:
prime#(0) → c (8)
prime#(s(0)) → c1 (9)
prime#(s(s(z0))) → c2(prime1#(s(s(z0)),s(z0))) (11)
prime1#(z0,0) → c3 (13)
prime1#(z0,s(0)) → c4 (15)
prime1#(z0,s(s(z1))) → c5(divp#(s(s(z1)),z0),prime1#(z0,s(z1))) (17)
divp#(z0,z1) → c6 (19)

1.1.1.1.1.1 Rule Shifting

The rules
divp#(z0,z1) → c6 (19)
are strictly oriented by the following linear polynomial interpretation over the naturals
[c] = 0
[c1] = 0
[c2(x1)] = 1 · x1 + 0
[c3] = 0
[c4] = 0
[c5(x1, x2)] = 1 · x1 + 0 + 1 · x2
[c6] = 0
[prime#(x1)] = 2 + 3 · x1
[prime1#(x1, x2)] = 1 · x2 + 0
[divp#(x1, x2)] = 1
[0] = 0
[s(x1)] = 2 + 1 · x1
which has the intended complexity. Here, only the following usable rules have been considered:
prime#(0) → c (8)
prime#(s(0)) → c1 (9)
prime#(s(s(z0))) → c2(prime1#(s(s(z0)),s(z0))) (11)
prime1#(z0,0) → c3 (13)
prime1#(z0,s(0)) → c4 (15)
prime1#(z0,s(s(z1))) → c5(divp#(s(s(z1)),z0),prime1#(z0,s(z1))) (17)
divp#(z0,z1) → c6 (19)

1.1.1.1.1.1.1 R is empty

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