Termination w.r.t. Q proof of /tmp/tmpFrC5FS/Ex1_Luc02b

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

    ↳5 QDP

        ↳6 QDPOrderProof (⇔)

        ↳7 QDP

        ↳8 DependencyGraphProof (⇔)

        ↳9 QDP

        ↳10 QDPOrderProof (⇔)

        ↳11 QDP

        ↳12 QDPOrderProof (⇔)

        ↳13 QDP

        ↳14 PisEmptyProof (⇔)

        ↳15 TRUE

    ↳16 QDP

        ↳17 QDPOrderProof (⇔)

        ↳18 QDP

        ↳19 PisEmptyProof (⇔)

        ↳20 TRUE

(0) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
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:

FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
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 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = ACTIVATE(x1)
n__from(x1) = x1
n__s(x1) = x1
n__first(x1, x2) = n__first(x1, x2)
FIRST(x1, x2) = FIRST(x2)
activate(x1) = x1
s(x1) = x1
cons(x1, x2) = x2
from(x1) = x1
first(x1, x2) = first(x1, x2)
0 = 0
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

[ACTIVATE₁, FIRST₁]
[n_{_first₂}, first₂] > nil

Status:

ACTIVATE₁: [1]
n_{_first₂}: [2,1]
FIRST₁: [1]
first₂: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
s(X) → n__s(X)
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
first(0, Z) → nil
first(X1, X2) → n__first(X1, X2)

(7) Obligation:

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(8) DependencyGraphProof (EQUIVALENT transformation)

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

(9) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__s(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__s(x1) = n__s(x1)
n__from(x1) = x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

n_{_s₁}: [1]

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

ACTIVATE(n__from(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__from(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__from(x1) = n__from(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

n_{_from₁}: [1]

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) TRUE

(16) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2) = SEL(x1, x2)
s(x1) = s(x1)
cons(x1, x2) = x2
activate(x1) = x1
n__from(x1) = n__from
from(x1) = from
n__s(x1) = n__s(x1)
n__first(x1, x2) = n__first(x1, x2)
first(x1, x2) = first(x1, x2)
0 = 0
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

[SEL₂, s₁, n_{_s₁}] > [n_{_first₂}, first₂] > nil
[n_{_from}, from]

Status:

SEL₂: [2,1]
s₁: [1]
n_{_from}: []
from: []
n_{_s₁}: [1]
n_{_first₂}: [2,1]
first₂: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
s(X) → n__s(X)
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
first(0, Z) → nil
first(X1, X2) → n__first(X1, X2)

(18) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(19) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) DependencyGraphProof (EQUIVALENT transformation)

(9) Obligation:

(10) QDPOrderProof (EQUIVALENT transformation)

(11) Obligation:

(12) QDPOrderProof (EQUIVALENT transformation)

(13) Obligation:

(14) PisEmptyProof (EQUIVALENT transformation)

(15) TRUE

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) TRUE