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 PisEmptyProof (⇔)
↳8 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)
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)  =  x1
h(x1)  =  h(x1)
f(x1)  =  f(x1)
a__f(x1)  =  a__f(x1)
a__c(x1)  =  a__c(x1)
g(x1)  =  g
d(x1)  =  d(x1)
a__h(x1)  =  a__h(x1)
mark(x1)  =  mark(x1)
c(x1)  =  c(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > [h1, ah1] > ac1 > [f1, d1, c1]
mark1 > af1 > ac1 > [f1, d1, c1]
mark1 > af1 > g > [f1, d1, c1]

Status:
h1: [1]
f1: [1]
af1: multiset
ac1: multiset
g: multiset
d1: multiset
ah1: [1]
mark1: [1]
c1: multiset


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:
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.

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE