Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHExIntrod_GM99

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
FILTER₂(mark₁(X1), X2) -> FILTER₂(X1, X2)
ACTIVE₁(divides₂(X1, X2)) -> ACTIVE₁(X2)
PROPER₁(filter₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(divides₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
PROPER₁(divides₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(head₁(X)) -> PROPER₁(X)
S₁(mark₁(X)) -> S₁(X)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> DIVIDES₂(s₁(s₁(X)), Y)
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
PROPER₁(divides₂(X1, X2)) -> DIVIDES₂(proper₁(X1), proper₁(X2))
ACTIVE₁(sieve₁(cons₂(X, Y))) -> CONS₂(X, filter₂(X, sieve₁(Y)))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
ACTIVE₁(tail₁(X)) -> TAIL₁(active₁(X))
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
HEAD₁(mark₁(X)) -> HEAD₁(X)
ACTIVE₁(filter₂(X1, X2)) -> FILTER₂(active₁(X1), X2)
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(divides₂(X1, X2)) -> DIVIDES₂(X1, active₁(X2))
PROPER₁(tail₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
DIVIDES₂(ok₁(X1), ok₁(X2)) -> DIVIDES₂(X1, X2)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(divides₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> TAIL₁(proper₁(X))
ACTIVE₁(primes) -> S₁(s₁(0))
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> SIEVE₁(Y)
ACTIVE₁(divides₂(X1, X2)) -> DIVIDES₂(active₁(X1), X2)
DIVIDES₂(mark₁(X1), X2) -> DIVIDES₂(X1, X2)
ACTIVE₁(primes) -> S₁(0)
HEAD₁(ok₁(X)) -> HEAD₁(X)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)
ACTIVE₁(sieve₁(X)) -> SIEVE₁(active₁(X))
TOP₁(mark₁(X)) -> PROPER₁(X)
FILTER₂(X1, mark₁(X2)) -> FILTER₂(X1, X2)
TAIL₁(ok₁(X)) -> TAIL₁(X)
PROPER₁(filter₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> CONS₂(Y, filter₂(X, sieve₁(Y)))
ACTIVE₁(primes) -> SIEVE₁(from₁(s₁(s₁(0))))
DIVIDES₂(X1, mark₁(X2)) -> DIVIDES₂(X1, X2)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
ACTIVE₁(sieve₁(cons₂(X, Y))) -> SIEVE₁(Y)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
PROPER₁(sieve₁(X)) -> SIEVE₁(proper₁(X))
ACTIVE₁(filter₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(filter₂(X1, X2)) -> FILTER₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
ACTIVE₁(filter₂(X1, X2)) -> ACTIVE₁(X2)
PROPER₁(sieve₁(X)) -> PROPER₁(X)
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(head₁(X)) -> HEAD₁(active₁(X))
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
SIEVE₁(ok₁(X)) -> SIEVE₁(X)
ACTIVE₁(filter₂(X1, X2)) -> FILTER₂(X1, active₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> IF₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
TAIL₁(mark₁(X)) -> TAIL₁(X)
PROPER₁(head₁(X)) -> HEAD₁(proper₁(X))
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
SIEVE₁(mark₁(X)) -> SIEVE₁(X)
FILTER₂(ok₁(X1), ok₁(X2)) -> FILTER₂(X1, X2)
ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(sieve₁(cons₂(X, Y))) -> FILTER₂(X, sieve₁(Y))
ACTIVE₁(from₁(X)) -> S₁(X)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> FILTER₂(s₁(s₁(X)), Z)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> FILTER₂(X, sieve₁(Y))
FROM₁(ok₁(X)) -> FROM₁(X)
ACTIVE₁(primes) -> FROM₁(s₁(s₁(0)))

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
FILTER₂(mark₁(X1), X2) -> FILTER₂(X1, X2)
ACTIVE₁(divides₂(X1, X2)) -> ACTIVE₁(X2)
PROPER₁(filter₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(divides₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
PROPER₁(divides₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(head₁(X)) -> PROPER₁(X)
S₁(mark₁(X)) -> S₁(X)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> DIVIDES₂(s₁(s₁(X)), Y)
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
PROPER₁(divides₂(X1, X2)) -> DIVIDES₂(proper₁(X1), proper₁(X2))
ACTIVE₁(sieve₁(cons₂(X, Y))) -> CONS₂(X, filter₂(X, sieve₁(Y)))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
ACTIVE₁(tail₁(X)) -> TAIL₁(active₁(X))
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
HEAD₁(mark₁(X)) -> HEAD₁(X)
ACTIVE₁(filter₂(X1, X2)) -> FILTER₂(active₁(X1), X2)
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(divides₂(X1, X2)) -> DIVIDES₂(X1, active₁(X2))
PROPER₁(tail₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
DIVIDES₂(ok₁(X1), ok₁(X2)) -> DIVIDES₂(X1, X2)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(divides₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> TAIL₁(proper₁(X))
ACTIVE₁(primes) -> S₁(s₁(0))
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> SIEVE₁(Y)
ACTIVE₁(divides₂(X1, X2)) -> DIVIDES₂(active₁(X1), X2)
DIVIDES₂(mark₁(X1), X2) -> DIVIDES₂(X1, X2)
ACTIVE₁(primes) -> S₁(0)
HEAD₁(ok₁(X)) -> HEAD₁(X)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)
ACTIVE₁(sieve₁(X)) -> SIEVE₁(active₁(X))
TOP₁(mark₁(X)) -> PROPER₁(X)
FILTER₂(X1, mark₁(X2)) -> FILTER₂(X1, X2)
TAIL₁(ok₁(X)) -> TAIL₁(X)
PROPER₁(filter₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> CONS₂(Y, filter₂(X, sieve₁(Y)))
ACTIVE₁(primes) -> SIEVE₁(from₁(s₁(s₁(0))))
DIVIDES₂(X1, mark₁(X2)) -> DIVIDES₂(X1, X2)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
ACTIVE₁(sieve₁(cons₂(X, Y))) -> SIEVE₁(Y)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
PROPER₁(sieve₁(X)) -> SIEVE₁(proper₁(X))
ACTIVE₁(filter₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(filter₂(X1, X2)) -> FILTER₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
ACTIVE₁(filter₂(X1, X2)) -> ACTIVE₁(X2)
PROPER₁(sieve₁(X)) -> PROPER₁(X)
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(head₁(X)) -> HEAD₁(active₁(X))
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
SIEVE₁(ok₁(X)) -> SIEVE₁(X)
ACTIVE₁(filter₂(X1, X2)) -> FILTER₂(X1, active₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> IF₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
TAIL₁(mark₁(X)) -> TAIL₁(X)
PROPER₁(head₁(X)) -> HEAD₁(proper₁(X))
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
SIEVE₁(mark₁(X)) -> SIEVE₁(X)
FILTER₂(ok₁(X1), ok₁(X2)) -> FILTER₂(X1, X2)
ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(sieve₁(cons₂(X, Y))) -> FILTER₂(X, sieve₁(Y))
ACTIVE₁(from₁(X)) -> S₁(X)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> FILTER₂(s₁(s₁(X)), Z)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> FILTER₂(X, sieve₁(Y))
FROM₁(ok₁(X)) -> FROM₁(X)
ACTIVE₁(primes) -> FROM₁(s₁(s₁(0)))

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 12 SCCs with 38 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DIVIDES₂(mark₁(X1), X2) -> DIVIDES₂(X1, X2)
DIVIDES₂(X1, mark₁(X2)) -> DIVIDES₂(X1, X2)
DIVIDES₂(ok₁(X1), ok₁(X2)) -> DIVIDES₂(X1, X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DIVIDES₂(ok₁(X1), ok₁(X2)) -> DIVIDES₂(X1, X2)

Used argument filtering: DIVIDES₂(x1, x2) = x2
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DIVIDES₂(mark₁(X1), X2) -> DIVIDES₂(X1, X2)
DIVIDES₂(X1, mark₁(X2)) -> DIVIDES₂(X1, X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DIVIDES₂(X1, mark₁(X2)) -> DIVIDES₂(X1, X2)

Used argument filtering: DIVIDES₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DIVIDES₂(mark₁(X1), X2) -> DIVIDES₂(X1, X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DIVIDES₂(mark₁(X1), X2) -> DIVIDES₂(X1, X2)

Used argument filtering: DIVIDES₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FILTER₂(ok₁(X1), ok₁(X2)) -> FILTER₂(X1, X2)
FILTER₂(mark₁(X1), X2) -> FILTER₂(X1, X2)
FILTER₂(X1, mark₁(X2)) -> FILTER₂(X1, X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FILTER₂(X1, mark₁(X2)) -> FILTER₂(X1, X2)

Used argument filtering: FILTER₂(x1, x2) = x2
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FILTER₂(ok₁(X1), ok₁(X2)) -> FILTER₂(X1, X2)
FILTER₂(mark₁(X1), X2) -> FILTER₂(X1, X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FILTER₂(ok₁(X1), ok₁(X2)) -> FILTER₂(X1, X2)

Used argument filtering: FILTER₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FILTER₂(mark₁(X1), X2) -> FILTER₂(X1, X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FILTER₂(mark₁(X1), X2) -> FILTER₂(X1, X2)

Used argument filtering: FILTER₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)

Used argument filtering: IF₃(x1, x2, x3) = x3
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

Used argument filtering: IF₃(x1, x2, x3) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TAIL₁(mark₁(X)) -> TAIL₁(X)
TAIL₁(ok₁(X)) -> TAIL₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TAIL₁(ok₁(X)) -> TAIL₁(X)

Used argument filtering: TAIL₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TAIL₁(mark₁(X)) -> TAIL₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TAIL₁(mark₁(X)) -> TAIL₁(X)

Used argument filtering: TAIL₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

HEAD₁(ok₁(X)) -> HEAD₁(X)
HEAD₁(mark₁(X)) -> HEAD₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

HEAD₁(mark₁(X)) -> HEAD₁(X)

Used argument filtering: HEAD₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

HEAD₁(ok₁(X)) -> HEAD₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

HEAD₁(ok₁(X)) -> HEAD₁(X)

Used argument filtering: HEAD₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

S₁(ok₁(X)) -> S₁(X)
S₁(mark₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(mark₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

S₁(ok₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(ok₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FROM₁(mark₁(X)) -> FROM₁(X)
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FROM₁(ok₁(X)) -> FROM₁(X)

Used argument filtering: FROM₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FROM₁(mark₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FROM₁(mark₁(X)) -> FROM₁(X)

Used argument filtering: FROM₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SIEVE₁(mark₁(X)) -> SIEVE₁(X)
SIEVE₁(ok₁(X)) -> SIEVE₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SIEVE₁(ok₁(X)) -> SIEVE₁(X)

Used argument filtering: SIEVE₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SIEVE₁(mark₁(X)) -> SIEVE₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SIEVE₁(mark₁(X)) -> SIEVE₁(X)

Used argument filtering: SIEVE₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(filter₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(filter₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(tail₁(X)) -> PROPER₁(X)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(divides₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(head₁(X)) -> PROPER₁(X)
PROPER₁(divides₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(sieve₁(X)) -> PROPER₁(X)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(filter₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(filter₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(divides₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(divides₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
filter₂(x1, x2) = filter₂(x1, x2)
tail₁(x1) = x1
if₃(x1, x2, x3) = if₃(x1, x2, x3)
divides₂(x1, x2) = divides₂(x1, x2)
head₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
sieve₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> PROPER₁(X)
PROPER₁(sieve₁(X)) -> PROPER₁(X)
PROPER₁(head₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(head₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
tail₁(x1) = x1
sieve₁(x1) = x1
head₁(x1) = head₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> PROPER₁(X)
PROPER₁(sieve₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(sieve₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
tail₁(x1) = x1
sieve₁(x1) = sieve₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(tail₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(tail₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
tail₁(x1) = tail₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(from₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = from₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

                                  ↳ QDP

                                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(filter₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(divides₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(divides₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(filter₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(filter₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(divides₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(divides₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(filter₂(X1, X2)) -> ACTIVE₁(X2)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
head₁(x1) = x1
filter₂(x1, x2) = filter₂(x1, x2)
divides₂(x1, x2) = divides₂(x1, x2)
cons₂(x1, x2) = x1
if₃(x1, x2, x3) = x1
from₁(x1) = x1
s₁(x1) = x1
tail₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
head₁(x1) = x1
cons₂(x1, x2) = x1
if₃(x1, x2, x3) = x1
from₁(x1) = x1
s₁(x1) = x1
tail₁(x1) = tail₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
head₁(x1) = x1
cons₂(x1, x2) = x1
if₃(x1, x2, x3) = x1
from₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(from₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
head₁(x1) = x1
cons₂(x1, x2) = x1
if₃(x1, x2, x3) = x1
from₁(x1) = from₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
head₁(x1) = x1
cons₂(x1, x2) = x1
if₃(x1, x2, x3) = if₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
head₁(x1) = x1
cons₂(x1, x2) = cons₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

                                  ↳ QDP

                                    ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(head₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
head₁(x1) = head₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

                                  ↳ QDP

                                    ↳ QDPAfsSolverProof

                                      ↳ QDP

                                        ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = sieve₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

                                  ↳ QDP

                                    ↳ QDPAfsSolverProof

                                      ↳ QDP

                                        ↳ QDPAfsSolverProof

                                          ↳ QDP

                                            ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The TRS R consists of the following rules:

active₁(primes) -> mark₁(sieve₁(from₁(s₁(s₁(0)))))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, Y))) -> mark₁(X)
active₁(tail₁(cons₂(X, Y))) -> mark₁(Y)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(filter₂(s₁(s₁(X)), cons₂(Y, Z))) -> mark₁(if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y)))))
active₁(sieve₁(cons₂(X, Y))) -> mark₁(cons₂(X, filter₂(X, sieve₁(Y))))
active₁(sieve₁(X)) -> sieve₁(active₁(X))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(head₁(X)) -> head₁(active₁(X))
active₁(tail₁(X)) -> tail₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(filter₂(X1, X2)) -> filter₂(active₁(X1), X2)
active₁(filter₂(X1, X2)) -> filter₂(X1, active₁(X2))
active₁(divides₂(X1, X2)) -> divides₂(active₁(X1), X2)
active₁(divides₂(X1, X2)) -> divides₂(X1, active₁(X2))
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
head₁(mark₁(X)) -> mark₁(head₁(X))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
filter₂(mark₁(X1), X2) -> mark₁(filter₂(X1, X2))
filter₂(X1, mark₁(X2)) -> mark₁(filter₂(X1, X2))
divides₂(mark₁(X1), X2) -> mark₁(divides₂(X1, X2))
divides₂(X1, mark₁(X2)) -> mark₁(divides₂(X1, X2))
proper₁(primes) -> ok₁(primes)
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(tail₁(X)) -> tail₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(filter₂(X1, X2)) -> filter₂(proper₁(X1), proper₁(X2))
proper₁(divides₂(X1, X2)) -> divides₂(proper₁(X1), proper₁(X2))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
head₁(ok₁(X)) -> ok₁(head₁(X))
tail₁(ok₁(X)) -> ok₁(tail₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
filter₂(ok₁(X1), ok₁(X2)) -> ok₁(filter₂(X1, X2))
divides₂(ok₁(X1), ok₁(X2)) -> ok₁(divides₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.