Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 PisEmptyProof (⇔)
↳10 TRUE

(0) Obligation:

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

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(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:

A__F(f(X)) → A__C(f(g(f(X))))
A__H(X) → A__C(d(X))
MARK(f(X)) → A__F(mark(X))
MARK(f(X)) → MARK(X)
MARK(c(X)) → A__C(X)
MARK(h(X)) → A__H(mark(X))
MARK(h(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(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 1 SCC with 5 less nodes.

(4) Obligation:

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

MARK(h(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(h(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
h(x1)  =  h(x1)
f(x1)  =  x1
a__f(x1)  =  x1
a__c(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1
a__h(x1)  =  a__h(x1)
mark(x1)  =  mark(x1)
c(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
mark1 > ah1 > h1

Status:
ah1: [1]
h1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

(6) Obligation:

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

MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(f(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
f(x1)  =  f(x1)
a__f(x1)  =  a__f(x1)
a__c(x1)  =  a__c
g(x1)  =  g
d(x1)  =  d
a__h(x1)  =  a__h
mark(x1)  =  mark(x1)
c(x1)  =  c
h(x1)  =  h

Lexicographic path order with status [LPO].
Precedence:
MARK1 > d
mark1 > af1 > f1 > d
mark1 > af1 > ac > c > d
mark1 > af1 > g > d
mark1 > ah > ac > c > d
mark1 > ah > h > d

Status:
c: []
MARK1: [1]
f1: [1]
ac: []
ah: []
af1: [1]
g: []
mark1: [1]
d: []
h: []

The following usable rules [FROCOS05] were oriented:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

(8) Obligation:

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

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE