Termination w.r.t. Q proof of Transformed_CSR_04_Ex9_BLR02

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

0 QTRS

↳1 DependencyPairsProof (⇔, 0 ms)

↳2 QDP

↳3 QDPOrderProof (⇔, 48 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 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:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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:

A__FILTER(cons(X, Y), s(N), M) → MARK(X)
A__SIEVE(cons(s(N), Y)) → MARK(N)
A__NATS(N) → MARK(N)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
A__ZPRIMES → A__NATS(s(s(0)))
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(filter(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2, X3)) → MARK(X2)
MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(nats(X)) → A__NATS(mark(X))
MARK(nats(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

(3) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(filter(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2, X3)) → MARK(X2)
MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → A__NATS(mark(X))
MARK(nats(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A__FILTER(x₁, ..., x₃) ) = 2x₁ + x₂ + 2

POL( A__NATS(x₁) ) = 2x₁ + 2

POL( A__SIEVE(x₁) ) = 2x₁

POL( a__nats(x₁) ) = 2x₁ + 1

POL( cons(x₁, x₂) ) = x₁ + 1

POL( mark(x₁) ) = x₁

POL( nats(x₁) ) = 2x₁ + 1

POL( s(x₁) ) = x₁

POL( filter(x₁, ..., x₃) ) = 2x₁ + x₂ + 2x₃ + 2

POL( a__filter(x₁, ..., x₃) ) = 2x₁ + x₂ + 2x₃ + 2

POL( sieve(x₁) ) = 2x₁

POL( a__sieve(x₁) ) = 2x₁

POL( zprimes ) = 2

POL( a__zprimes ) = 2

POL( 0 ) = 0

POL( MARK(x₁) ) = x₁ + 2

POL( A__ZPRIMES ) = 2

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

a__nats(N) → cons(mark(N), nats(s(N)))
a__nats(X) → nats(X)
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(X) → sieve(X)
a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__zprimes → zprimes
a__zprimes → a__sieve(a__nats(s(s(0))))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))

(4) Obligation:

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

A__SIEVE(cons(s(N), Y)) → MARK(N)
A__NATS(N) → MARK(N)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
A__ZPRIMES → A__NATS(s(s(0)))
MARK(sieve(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(6) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(sieve(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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:

MARK(s(X)) → MARK(X)
MARK(sieve(X)) → MARK(X)

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:

MARK(s(X)) → MARK(X)
The graph contains the following edges 1 > 1
MARK(sieve(X)) → MARK(X)
The graph contains the following edges 1 > 1

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) QDPOrderProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Obligation:

(7) UsableRulesProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPSizeChangeProof (EQUIVALENT transformation)

(10) YES