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 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 QDPOrderProof (⇔)
    ↳29 QDP
    ↳30 QDPOrderProof (⇔)
    ↳31 QDP
    ↳32 PisEmptyProof (⇔)
    ↳33 TRUE
  ↳34 QDP
    ↳35 QDPOrderProof (⇔)
    ↳36 QDP
    ↳37 PisEmptyProof (⇔)
    ↳38 TRUE

(0) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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(and(tt, X)) → MARK(X)
ACTIVE(plus(N, 0)) → MARK(N)
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
ACTIVE(plus(N, s(M))) → S(plus(N, M))
ACTIVE(plus(N, s(M))) → PLUS(N, M)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(tt) → ACTIVE(tt)
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(plus(X1, X2)) → PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(active(X1), X2) → PLUS(X1, X2)
PLUS(X1, active(X2)) → PLUS(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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 4 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) 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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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

(7) 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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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

(9) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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:

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

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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.


PLUS(X1, mark(X2)) → PLUS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(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

(14) Obligation:

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

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

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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.


PLUS(mark(X1), X2) → PLUS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
mark1 > PLUS2

Status:
PLUS2: [1,2]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

PLUS(active(X1), X2) → PLUS(X1, X2)
PLUS(X1, active(X2)) → PLUS(X1, X2)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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.


PLUS(X1, active(X2)) → PLUS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(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

(18) Obligation:

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

PLUS(active(X1), X2) → PLUS(X1, X2)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(active(X1), X2) → PLUS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(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

(20) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

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

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


AND(X1, mark(X2)) → AND(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(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

(25) Obligation:

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

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

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


AND(mark(X1), X2) → AND(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2)  =  AND(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
mark1 > AND2

Status:
AND2: [1,2]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


AND(X1, active(X2)) → AND(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(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

(29) Obligation:

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

AND(active(X1), X2) → AND(X1, X2)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


AND(active(X1), X2) → AND(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(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

(31) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(and(tt, X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
ACTIVE(plus(N, 0)) → MARK(N)
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(and(tt, X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
ACTIVE(plus(N, 0)) → MARK(N)
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
and(x1, x2)  =  and(x1, x2)
ACTIVE(x1)  =  x1
mark(x1)  =  x1
tt  =  tt
plus(x1, x2)  =  plus(x1, x2)
0  =  0
s(x1)  =  s(x1)
active(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
and2 > MARK1
plus2 > MARK1
plus2 > s1
0 > MARK1

Status:
plus2: [1,2]
MARK1: [1]
tt: []
s1: [1]
and2: [2,1]
0: []

The following usable rules [FROCOS05] were oriented:

mark(tt) → active(tt)
mark(s(X)) → active(s(mark(X)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
active(and(tt, X)) → mark(X)
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)

(36) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(37) PisEmptyProof (EQUIVALENT transformation)

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

(38) TRUE