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 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE
    ↳20 QDP
        ↳21 QDPOrderProof (⇔)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔)
        ↳24 QDP
        ↳25 DependencyGraphProof (⇔)
        ↳26 TRUE

(0) Obligation:

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

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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(X)) → MARK(g(h(f(X))))
ACTIVE(f(X)) → G(h(f(X)))
ACTIVE(f(X)) → H(f(X))
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → F(mark(X))
MARK(f(X)) → MARK(X)
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → H(mark(X))
MARK(h(X)) → MARK(X)
F(mark(X)) → F(X)
F(active(X)) → F(X)
G(mark(X)) → G(X)
G(active(X)) → G(X)
H(mark(X)) → H(X)
H(active(X)) → H(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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.


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

POL(H(x1)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(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(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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:

G(active(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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.


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

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(F(x1)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(f(X)) → MARK(X)
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → MARK(X)
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = 1   
POL(MARK(x1)) = 1 + x1   
POL(active(x1)) = 0   
POL(f(x1)) = 1 + x1   
POL(g(x1)) = 0   
POL(h(x1)) = 1 + x1   
POL(mark(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

g(active(X)) → g(X)
g(mark(X)) → g(X)

(22) Obligation:

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

ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(g(X)) → ACTIVE(g(X))

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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(f(X)) → MARK(g(h(f(X))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 0   
POL(active(x1)) = 0   
POL(f(x1)) = 1   
POL(g(x1)) = 0   
POL(h(x1)) = 0   
POL(mark(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

g(active(X)) → g(X)
g(mark(X)) → g(X)

(24) Obligation:

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

MARK(g(X)) → ACTIVE(g(X))

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(26) TRUE