YES Termination w.r.t. Q proof of Transformed_CSR_04_Ex9_BLR02_Z.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 50 ms)
↳6 QDP
↳7 UsableRulesProof (⇔, 0 ms)
↳8 QDP
↳9 QDPSizeChangeProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

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

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(s(s(0)))
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
ACTIVATE(n__sieve(X)) → SIEVE(X)
ACTIVATE(n__nats(X)) → NATS(X)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

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

ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ACTIVATE(n__sieve(X)) → SIEVE(X)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__sieve(X)) → SIEVE(X)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( FILTER(x1, ..., x3) ) = 2x1

POL( activate(x1) ) = x1 + 1

POL( n__filter(x1, ..., x3) ) = max{0, x1 - 1}

POL( filter(x1, ..., x3) ) = x1

POL( n__sieve(x1) ) = 2x1 + 1

POL( sieve(x1) ) = 2x1 + 2

POL( n__nats(x1) ) = 0

POL( nats(x1) ) = 1

POL( cons(x1, x2) ) = x2 + 1

POL( 0 ) = 1

POL( s(x1) ) = 1

POL( ACTIVATE(x1) ) = 2x1 + 2

POL( SIEVE(x1) ) = 2x1 + 2


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X
filter(X1, X2, X3) → n__filter(X1, X2, X3)
filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
sieve(X) → n__sieve(X)
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
nats(N) → cons(N, n__nats(s(N)))
nats(X) → n__nats(X)

(6) Obligation:

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

ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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

(7) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(8) Obligation:

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

ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
    The graph contains the following edges 1 > 1, 1 > 2, 1 > 3

  • FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
    The graph contains the following edges 1 > 1

  • FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
    The graph contains the following edges 1 > 1

(10) YES