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(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
ACTIVE(f(g(X), Y)) → F(X, f(g(X), Y))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(f(X1, X2)) → F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → G(mark(X))
MARK(g(X)) → MARK(X)
F(mark(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
G(mark(X)) → G(X)
G(active(X)) → G(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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 3 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(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(X1, mark(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(X1, mark(X2)) → F(X1, X2) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

  • F(mark(X1), X2) → F(X1, X2) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

  • F(active(X1), X2) → F(X1, X2) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

  • F(X1, active(X2)) → F(X1, X2) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

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:

MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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 together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation [POLO]:


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

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

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

  • MARK(f(X1, X2)) → MARK(X1) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

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

  • MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 >= 1

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

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


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

(13) TRUE