Termination w.r.t. Q proof of /tmp/tmpEoQk6c/ExIntrod_GM01

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 PisEmptyProof (⇔)

    ↳32 TRUE

  ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 PisEmptyProof (⇔)

    ↳43 TRUE

  ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 PisEmptyProof (⇔)

    ↳50 TRUE

  ↳51 QDP

(0) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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:

ACTIVE(incr(nil)) → MARK(nil)
ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
ACTIVE(incr(cons(X, L))) → CONS(s(X), incr(L))
ACTIVE(incr(cons(X, L))) → S(X)
ACTIVE(incr(cons(X, L))) → INCR(L)
ACTIVE(adx(nil)) → MARK(nil)
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
ACTIVE(adx(cons(X, L))) → INCR(cons(X, adx(L)))
ACTIVE(adx(cons(X, L))) → CONS(X, adx(L))
ACTIVE(adx(cons(X, L))) → ADX(L)
ACTIVE(nats) → MARK(adx(zeros))
ACTIVE(nats) → ADX(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(head(cons(X, L))) → MARK(X)
ACTIVE(tail(cons(X, L))) → MARK(L)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(nil) → ACTIVE(nil)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(adx(X)) → ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(nats) → ACTIVE(nats)
MARK(zeros) → ACTIVE(zeros)
MARK(0) → ACTIVE(0)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → HEAD(mark(X))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → TAIL(mark(X))
MARK(tail(X)) → MARK(X)
INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
ADX(mark(X)) → ADX(X)
ADX(active(X)) → ADX(X)
HEAD(mark(X)) → HEAD(X)
HEAD(active(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
TAIL(active(X)) → TAIL(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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 7 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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.

TAIL(mark(X)) → TAIL(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > TAIL₁

Status:

mark₁: [1]
TAIL₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

TAIL(active(X)) → TAIL(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAIL(active(X)) → TAIL(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

HEAD(active(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

HEAD(mark(X)) → HEAD(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > HEAD₁

Status:

mark₁: [1]
HEAD₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

HEAD(active(X)) → HEAD(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

HEAD(active(X)) → HEAD(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

ADX(active(X)) → ADX(X)
ADX(mark(X)) → ADX(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADX(mark(X)) → ADX(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > ADX₁

Status:

ADX₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

ADX(active(X)) → ADX(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADX(active(X)) → ADX(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > S₁

Status:

mark₁: [1]
S₁: [1]

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

S(active(X)) → S(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, mark(X2)) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > CONS₂

Status:

CONS₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

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

CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, active(X2)) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(active(X1), X2) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

INCR(active(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INCR(mark(X)) → INCR(X)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > INCR₁

Status:

mark₁: [1]
INCR₁: [1]

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

INCR(active(X)) → INCR(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INCR(active(X)) → INCR(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE

(51) Obligation:

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

ACTIVE(incr(cons(X, L))) → MARK(cons(s(X), incr(L)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(adx(cons(X, L))) → MARK(incr(cons(X, adx(L))))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(head(cons(X, L))) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(tail(cons(X, L))) → MARK(L)
MARK(s(X)) → MARK(X)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(adx(X)) → MARK(X)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
mark(incr(X)) → active(incr(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(adx(X)) → active(adx(mark(X)))
mark(nats) → active(nats)
mark(zeros) → active(zeros)
mark(0) → active(0)
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
adx(mark(X)) → adx(X)
adx(active(X)) → adx(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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