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 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE
    ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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 3 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) 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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)

From the DPs we obtained the following set of size-change graphs:

  • G(active(X)) → G(X) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • G(mark(X)) → G(X) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)

From the DPs we obtained the following set of size-change graphs:

  • F(active(X)) → F(X) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • F(mark(X)) → F(X) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(11) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Combined order from the following AFS and order.
g(x1)  =  g
active(x1)  =  x1
mark(x1)  =  mark(x1)
f(x1)  =  f(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

f1 > [g, mark1]

Status:
g: []
mark1: [1]
f1: [1]

AFS:
g(x1)  =  g
active(x1)  =  x1
mark(x1)  =  mark(x1)
f(x1)  =  f(x1)

From the DPs we obtained the following set of size-change graphs:

  • ACTIVE(f(g(X))) → MARK(g(X)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • MARK(f(X)) → ACTIVE(f(X)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 >= 1

  • MARK(g(X)) → ACTIVE(g(X)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 >= 1

We oriented the following set of usable rules [AAECC05,FROCOS05].


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

(13) TRUE