Termination w.r.t. Q proof of /tmp/tmpdtc_A9/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 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)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
A__FIRST(x0, x1, x2) = A__FIRST(x2)
MARK(x0, x1) = MARK(x0, x1)
A__FROM(x0, x1) = A__FROM(x0, x1)

Tags:
A__FIRST has argument tags [0,0,6] and root tag 3
MARK has argument tags [6,0] and root tag 0
A__FROM has argument tags [1,6] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
A__FIRST(x1, x2) = x1
s(x1) = x1
cons(x1, x2) = x1
MARK(x1) = x1
A__FROM(x1) = A__FROM
first(x1, x2) = first(x1, x2)
mark(x1) = x1
from(x1) = x1
a__first(x1, x2) = a__first(x1, x2)
a__from(x1) = x1
0 = 0
nil = nil

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

[first₂, a_{_first₂}, nil] > A_{_FROM}
0 > A_{_FROM}

Status:

A_{_FROM}: []
first₂: [2,1]
a_{_first₂}: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

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

(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(from(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
A__FROM(x0, x1) = A__FROM(x0, x1)
MARK(x0, x1) = MARK(x1)

Tags:
A__FROM has argument tags [0,2] and root tag 0
MARK has argument tags [3,2] and root tag 0

Comparison: MIN
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
A__FROM(x1) = A__FROM(x1)
MARK(x1) = MARK
from(x1) = from(x1)
mark(x1) = mark(x1)
s(x1) = x1
cons(x1, x2) = x1
first(x1, x2) = first(x1, x2)
a__first(x1, x2) = a__first(x1, x2)
a__from(x1) = a__from(x1)
0 = 0
nil = nil

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

[from₁, mark₁, first₂, a_{_first₂}, a_{_from₁}, nil] > A_{_FROM₁} > MARK
[from₁, mark₁, first₂, a_{_first₂}, a_{_from₁}, nil] > 0

Status:

A_{_FROM₁}: [1]
MARK: []
from₁: [1]
mark₁: [1]
first₂: [2,1]
a_{_first₂}: [2,1]
a_{_from₁}: [1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

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

(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(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.

(7) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
A__FROM(x0, x1) = A__FROM(x0, x1)
MARK(x0, x1) = MARK(x1)

Tags:
A__FROM has argument tags [0,1] and root tag 0
MARK has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
A__FROM(x1) = x1
MARK(x1) = MARK
from(x1) = from(x1)
mark(x1) = mark(x1)
s(x1) = s(x1)
cons(x1, x2) = x1
first(x1, x2) = first(x2)
a__first(x1, x2) = a__first(x2)
a__from(x1) = a__from(x1)
0 = 0
nil = nil

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

MARK > [from₁, mark₁, s₁, first₁, a_{_first₁}, a_{_from₁}, nil]
0 > [from₁, mark₁, s₁, first₁, a_{_first₁}, a_{_from₁}, nil]

Status:

MARK: []
from₁: [1]
mark₁: [1]
s₁: [1]
first₁: [1]
a_{_first₁}: [1]
a_{_from₁}: [1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

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

(8) 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.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
A__FROM(x0, x1) = A__FROM(x0, x1)
MARK(x0, x1) = MARK(x1)

Tags:
A__FROM has argument tags [3,1] and root tag 0
MARK has argument tags [0,3] and root tag 1

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
A__FROM(x1) = A__FROM(x1)
MARK(x1) = MARK
from(x1) = from(x1)
mark(x1) = x1
cons(x1, x2) = cons(x1)
first(x1, x2) = x2
a__first(x1, x2) = x2
a__from(x1) = a__from(x1)
0 = 0
nil = nil
s(x1) = x1

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

[A_{_FROM₁}, from₁, cons₁, a_{_from₁}] > [MARK, nil]
0 > [MARK, nil]

Status:

A_{_FROM₁}: [1]
MARK: []
from₁: [1]
cons₁: [1]
a_{_from₁}: [1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

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

(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) QDPOrderProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) PisEmptyProof (EQUIVALENT transformation)

(12) TRUE