Termination w.r.t. Q proof of /tmp/tmpZ1TGos/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 QDPOrderProof (⇔)

↳8 QDP

↳9 DependencyGraphProof (⇔)

↳10 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)

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) = x1
cons(x1, x2) = x1
MARK(x1) = MARK(x1)
A__FROM(x1) = A__FROM(x1)
first(x1, x2) = first(x1, x2)
mark(x1) = x1
from(x1) = x1
a__from(x1) = x1
a__first(x1, x2) = a__first(x1, x2)
nil = nil
0 = 0

Recursive Path Order [RPO].
Precedence:

[A_{_FIRST₁}, first₂, a_{_first₂}] > [MARK₁, A_{_FROM₁}] > nil
0 > nil

The following usable rules [FROCOS05] were oriented:

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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__FROM(x1) = x1
MARK(x1) = x1
from(x1) = x1
mark(x1) = x1
s(x1) = s(x1)
cons(x1, x2) = x1
a__first(x1, x2) = a__first(x2)
first(x1, x2) = first(x2)
nil = nil
a__from(x1) = x1
0 = 0

Recursive Path Order [RPO].
Precedence:

s₁ > [a_{_first₁}, first₁] > nil
0 > nil

The following usable rules [FROCOS05] were oriented:

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

(6) 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(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) 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))
MARK(from(X)) → MARK(X)
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.
A__FROM(x1) = x1
MARK(x1) = x1
from(x1) = from(x1)
mark(x1) = x1
cons(x1, x2) = cons(x1)
a__first(x1, x2) = x2
first(x1, x2) = x2
nil = nil
s(x1) = x1
a__from(x1) = a__from(x1)
0 = 0

Recursive Path Order [RPO].
Precedence:

0 > [from₁, cons₁, nil, a_{_from₁}]

The following usable rules [FROCOS05] were oriented:

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

(8) Obligation:

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

A__FROM(X) → MARK(X)

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) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) QDPOrderProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) QDPOrderProof (EQUIVALENT transformation)

(8) Obligation:

(9) DependencyGraphProof (EQUIVALENT transformation)

(10) TRUE