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

(0) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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(f(g(X), Y)) → F(X, f(g(X), Y))
ACTIVE(f(X1, X2)) → F(active(X1), X2)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
ACTIVE(g(X)) → G(active(X))
ACTIVE(g(X)) → ACTIVE(X)
F(mark(X1), X2) → F(X1, X2)
G(mark(X)) → G(X)
PROPER(f(X1, X2)) → F(proper(X1), proper(X2))
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(f(X1, X2)) → PROPER(X2)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
F(ok(X1), ok(X2)) → F(X1, X2)
G(ok(X)) → G(X)
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(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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 5 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

G(ok(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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.


G(ok(X)) → G(X)
G(mark(X)) → G(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(G(x1)) = x1   
POL(mark(x1)) = 1 + x1   
POL(ok(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

F(ok(X1), ok(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(ok(X1), ok(X2)) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F(x1, x2)) = x2   
POL(mark(x1)) = 0   
POL(ok(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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.


F(mark(X1), X2) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F(x1, x2)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

PROPER(f(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(f(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(g(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(PROPER(x1)) = x1   
POL(f(x1, x2)) = 1 + x1 + x2   
POL(g(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(f(x1, x2)) = 1 + x1   
POL(g(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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.

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) 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(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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.

(28) 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))
TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TOP(x1)) = x1   
POL(active(x1)) = x1   
POL(f(x1, x2)) = 1 + x1   
POL(g(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   
POL(ok(x1)) = 1 + x1   
POL(proper(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
g(mark(X)) → mark(g(X))
g(ok(X)) → ok(g(X))
f(mark(X1), X2) → mark(f(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))

(29) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
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.

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE