(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)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
and(x1, x2)  =  and(x1, x2)
tt  =  tt
plus(x1, x2)  =  plus(x1)
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
tt > [mark1, and2, plus1] > 0 > s

Status:
mark1: [1]
and2: [2,1]
tt: []
plus1: [1]
0: []
s: []


The following usable rules [FROCOS05] were oriented:

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)

(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)  =  S(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x2)
tt  =  tt
mark(x1)  =  mark(x1)
plus(x1, x2)  =  plus(x1, x2)
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
S1 > s
and1 > [mark1, plus2, 0] > active1 > s
and1 > [mark1, plus2, 0] > tt > s

Status:
S1: [1]
active1: [1]
and1: [1]
tt: []
mark1: [1]
plus2: [1,2]
0: []
s: []


The following usable rules [FROCOS05] were oriented:

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)

(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(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)
mark(x1)  =  mark(x1)
active(x1)  =  x1
and(x1, x2)  =  and(x2)
tt  =  tt
plus(x1, x2)  =  plus(x1)
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
PLUS1 > [mark1, plus1, s]
and1 > [mark1, plus1, s]
tt > [mark1, plus1, s]
0 > [mark1, plus1, s]

Status:
PLUS1: [1]
mark1: [1]
and1: [1]
tt: []
plus1: [1]
0: []
s: []


The following usable rules [FROCOS05] were oriented:

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)

(14) Obligation:

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

PLUS(X1, mark(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(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)  =  PLUS(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x2)
tt  =  tt
plus(x1, x2)  =  plus(x1, x2)
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
and1 > [mark1, plus2] > PLUS1 > [active1, tt, s]
and1 > [mark1, plus2] > 0 > [active1, tt, s]

Status:
PLUS1: [1]
mark1: [1]
active1: [1]
and1: [1]
tt: []
plus2: [1,2]
0: []
s: []


The following usable rules [FROCOS05] were oriented:

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)

(16) Obligation:

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

PLUS(X1, mark(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, mark(X2)) → PLUS(X1, X2)
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)  =  PLUS(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x2)
tt  =  tt
plus(x1, x2)  =  plus(x1, x2)
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
PLUS2 > [active1, 0, s]
and1 > [mark1, plus2] > [active1, 0, s]
tt > [mark1, plus2] > [active1, 0, s]

Status:
PLUS2: [2,1]
mark1: [1]
active1: [1]
and1: [1]
tt: []
plus2: [1,2]
0: []
s: []


The following usable rules [FROCOS05] were oriented:

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)

(18) 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.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) 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.

(22) 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)
mark(x1)  =  mark(x1)
active(x1)  =  x1
and(x1, x2)  =  and(x2)
tt  =  tt
plus(x1, x2)  =  plus(x1)
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
AND1 > [mark1, plus1, s]
and1 > [mark1, plus1, s]
tt > [mark1, plus1, s]
0 > [mark1, plus1, s]

Status:
AND1: [1]
mark1: [1]
and1: [1]
tt: []
plus1: [1]
0: []
s: []


The following usable rules [FROCOS05] were oriented:

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)

(23) Obligation:

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

AND(X1, mark(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(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)  =  AND(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x2)
tt  =  tt
plus(x1, x2)  =  plus(x1, x2)
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
and1 > [mark1, plus2] > AND1 > [active1, tt, s]
and1 > [mark1, plus2] > 0 > [active1, tt, s]

Status:
AND1: [1]
mark1: [1]
active1: [1]
and1: [1]
tt: []
plus2: [1,2]
0: []
s: []


The following usable rules [FROCOS05] were oriented:

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)

(25) Obligation:

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

AND(X1, mark(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(X1, mark(X2)) → AND(X1, X2)
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)  =  AND(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
and(x1, x2)  =  and(x2)
tt  =  tt
plus(x1, x2)  =  plus(x1, x2)
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:
AND2 > [active1, 0, s]
and1 > [mark1, plus2] > [active1, 0, s]
tt > [mark1, plus2] > [active1, 0, s]

Status:
AND2: [2,1]
mark1: [1]
active1: [1]
and1: [1]
tt: []
plus2: [1,2]
0: []
s: []


The following usable rules [FROCOS05] were oriented:

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)

(27) 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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) 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.

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(and(tt, X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(plus(N, 0)) → MARK(N)
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
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)  =  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].
Quasi-Precedence:
plus2 > s1

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


The following usable rules [FROCOS05] were oriented:

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)

(32) Obligation:

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

MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(s(X)) → ACTIVE(s(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.

(33) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(34) TRUE