Termination w.r.t. Q proof of /tmp/tmpbZLhnk/Ex6_Luc98

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 QDPOrderProof (⇔)

↳4 QDP

↳5 QDPOrderProof (⇔)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 QDP

↳9 QDPOrderProof (⇔)

↳10 QDP

↳11 PisEmptyProof (⇔)

↳12 TRUE

(0) Obligation:

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

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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:

A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
A__FROM(X) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(from(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__FIRST(x1, x2) = A__FIRST(x2)
s(x1) = s(x1)
cons(x1, x2) = x1
MARK(x1) = x1
A__FROM(x1) = x1
first(x1, x2) = first(x1, x2)
mark(x1) = mark(x1)
from(x1) = from(x1)
a__from(x1) = a__from(x1)
a__first(x1, x2) = a__first(x1, x2)
nil = nil
0 = 0

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

[first₂, mark₁, from₁, a_{_from₁}, a_{_first₂}] > A_{_FIRST₁}
[first₂, mark₁, from₁, a_{_from₁}, a_{_first₂}] > s₁
[first₂, mark₁, from₁, a_{_from₁}, a_{_first₂}] > 0 > nil

Status:

a_{_first₂}: [2,1]
from₁: [1]
A_{_FIRST₁}: [1]
mark₁: [1]
s₁: [1]
a_{_from₁}: [1]
first₂: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

a__from(X) → from(X)
a__first(X1, X2) → first(X1, X2)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
mark(from(X)) → a__from(mark(X))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))

(4) Obligation:

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

A__FROM(X) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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].

The following pairs can be oriented strictly and are deleted.

MARK(from(X)) → A__FROM(mark(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__FROM(x1) = x1
MARK(x1) = x1
from(x1) = from(x1)
mark(x1) = x1
cons(x1, x2) = x1
a__from(x1) = a__from(x1)
s(x1) = s(x1)
first(x1, x2) = x2
a__first(x1, x2) = x2
0 = 0
nil = nil

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

[from₁, a_{_from₁}] > [0, nil]
s₁ > [0, nil]

Status:

from₁: [1]
a_{_from₁}: [1]
s₁: [1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → from(X)
a__first(X1, X2) → first(X1, X2)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
mark(from(X)) → a__from(mark(X))

(6) Obligation:

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

A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
cons(x1, x2) = cons(x1, x2)

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

trivial

Status:

cons₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

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

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

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

(11) 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) QDPOrderProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) PisEmptyProof (EQUIVALENT transformation)

(12) TRUE