### (0) Obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(1, 1).

The TRS R consists of the following rules:

filter(cons(X), 0, M) → cons(0)
filter(cons(X), s(N), M) → cons(X)
sieve(cons(0)) → cons(0)
sieve(cons(s(N))) → cons(s(N))
nats(N) → cons(N)
zprimessieve(nats(s(s(0))))

Rewrite Strategy: FULL

### (1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

As the TRS is a non-duplicating overlay system, we have rc = irc.

### (2) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, 1).

The TRS R consists of the following rules:

filter(cons(X), 0, M) → cons(0)
filter(cons(X), s(N), M) → cons(X)
sieve(cons(0)) → cons(0)
sieve(cons(s(N))) → cons(s(N))
nats(N) → cons(N)
zprimessieve(nats(s(s(0))))

Rewrite Strategy: INNERMOST

### (3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

filter(cons(z0), 0, z1) → cons(0)
filter(cons(z0), s(z1), z2) → cons(z0)
sieve(cons(0)) → cons(0)
sieve(cons(s(z0))) → cons(s(z0))
nats(z0) → cons(z0)
zprimessieve(nats(s(s(0))))
Tuples:

FILTER(cons(z0), 0, z1) → c
FILTER(cons(z0), s(z1), z2) → c1
SIEVE(cons(0)) → c2
SIEVE(cons(s(z0))) → c3
NATS(z0) → c4
ZPRIMESc5(SIEVE(nats(s(s(0)))), NATS(s(s(0))))
S tuples:

FILTER(cons(z0), 0, z1) → c
FILTER(cons(z0), s(z1), z2) → c1
SIEVE(cons(0)) → c2
SIEVE(cons(s(z0))) → c3
NATS(z0) → c4
ZPRIMESc5(SIEVE(nats(s(s(0)))), NATS(s(s(0))))
K tuples:none
Defined Rule Symbols:

filter, sieve, nats, zprimes

Defined Pair Symbols:

FILTER, SIEVE, NATS, ZPRIMES

Compound Symbols:

c, c1, c2, c3, c4, c5

### (5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 6 trailing nodes:

FILTER(cons(z0), s(z1), z2) → c1
FILTER(cons(z0), 0, z1) → c
SIEVE(cons(s(z0))) → c3
NATS(z0) → c4
SIEVE(cons(0)) → c2
ZPRIMESc5(SIEVE(nats(s(s(0)))), NATS(s(s(0))))

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

filter(cons(z0), 0, z1) → cons(0)
filter(cons(z0), s(z1), z2) → cons(z0)
sieve(cons(0)) → cons(0)
sieve(cons(s(z0))) → cons(s(z0))
nats(z0) → cons(z0)
zprimessieve(nats(s(s(0))))
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:

filter, sieve, nats, zprimes

Defined Pair Symbols:none

Compound Symbols:none

### (7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty