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 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 PisEmptyProof (⇔)
    ↳39 TRUE
  ↳40 QDP
    ↳41 QDPOrderProof (⇔)
    ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 PisEmptyProof (⇔)
    ↳46 TRUE
  ↳47 QDP
    ↳48 QDPOrderProof (⇔)
    ↳49 QDP
    ↳50 QDPOrderProof (⇔)
    ↳51 QDP
    ↳52 PisEmptyProof (⇔)
    ↳53 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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(plus(N, s(M))) → S(plus(N, M))
ACTIVE(plus(N, s(M))) → PLUS(N, M)
ACTIVE(x(N, s(M))) → PLUS(x(N, M), N)
ACTIVE(x(N, s(M))) → X(N, M)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → PLUS(active(X1), X2)
ACTIVE(plus(X1, X2)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → PLUS(X1, active(X2))
ACTIVE(plus(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(x(X1, X2)) → X(active(X1), X2)
ACTIVE(x(X1, X2)) → ACTIVE(X1)
ACTIVE(x(X1, X2)) → X(X1, active(X2))
ACTIVE(x(X1, X2)) → ACTIVE(X2)
AND(mark(X1), X2) → AND(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
S(mark(X)) → S(X)
X(mark(X1), X2) → X(X1, X2)
X(X1, mark(X2)) → X(X1, X2)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(plus(X1, X2)) → PLUS(proper(X1), proper(X2))
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(x(X1, X2)) → X(proper(X1), proper(X2))
PROPER(x(X1, X2)) → PROPER(X1)
PROPER(x(X1, X2)) → PROPER(X2)
AND(ok(X1), ok(X2)) → AND(X1, X2)
PLUS(ok(X1), ok(X2)) → PLUS(X1, X2)
S(ok(X)) → S(X)
X(ok(X1), ok(X2)) → X(X1, X2)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 16 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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


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

Lexicographic Path Order [LPO].
Precedence:
trivial


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

S(ok(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


S(ok(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
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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


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
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


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

PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(ok(X1), ok(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(ok(X1), ok(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
mark(x1)  =  mark(x1)
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

PLUS(X1, mark(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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)

Lexicographic Path Order [LPO].
Precedence:
trivial


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

AND(ok(X1), ok(X2)) → AND(X1, X2)
AND(mark(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

AND(ok(X1), ok(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


AND(ok(X1), ok(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(x2)
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:
ok1 > AND1


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

PROPER(and(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(x(X1, X2)) → PROPER(X1)
PROPER(x(X1, X2)) → PROPER(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

PROPER(s(X)) → PROPER(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(37) 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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) Obligation:

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

ACTIVE(plus(X1, X2)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(x(X1, X2)) → ACTIVE(X1)
ACTIVE(x(X1, X2)) → ACTIVE(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

ACTIVE(s(X)) → ACTIVE(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(44) 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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
proper(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
tt  =  tt
plus(x1, x2)  =  plus(x1, x2)
0  =  0
s(x1)  =  s(x1)
x(x1, x2)  =  x(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
and2 > [TOP1, mark1, tt, 0]
x2 > plus2 > s1 > [TOP1, mark1, tt, 0]


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

(49) Obligation:

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

TOP(ok(X)) → TOP(active(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(ok(X)) → TOP(active(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
and(x1, x2)  =  x2
tt  =  tt
mark(x1)  =  mark
plus(x1, x2)  =  x1
0  =  0
s(x1)  =  s(x1)
x(x1, x2)  =  x1

Lexicographic Path Order [LPO].
Precedence:
tt > [TOP1, mark, 0]
s1 > ok1 > [TOP1, mark, 0]


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

(51) 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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
active(x(X1, X2)) → x(active(X1), X2)
active(x(X1, X2)) → x(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
x(mark(X1), X2) → mark(x(X1, X2))
x(X1, mark(X2)) → mark(x(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(x(X1, X2)) → x(proper(X1), proper(X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
x(ok(X1), ok(X2)) → ok(x(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(52) PisEmptyProof (EQUIVALENT transformation)

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

(53) TRUE