(0) Obligation:

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

a__g(X) → a__h(X)
a__cd
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__cc

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__G(X) → A__H(X)
A__H(d) → A__G(c)
MARK(g(X)) → A__G(X)
MARK(h(X)) → A__H(X)
MARK(c) → A__C

The TRS R consists of the following rules:

a__g(X) → a__h(X)
a__cd
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__cc

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 3 less nodes.

(4) Obligation:

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

A__H(d) → A__G(c)
A__G(X) → A__H(X)

The TRS R consists of the following rules:

a__g(X) → a__h(X)
a__cd
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__cc

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.


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

Recursive Path Order [RPO].
Precedence:
[ac, mark1] > d > c > [AH1, AG1]
[ac, mark1] > [ag, ah, h] > c > [AH1, AG1]
[ac, mark1] > [ag, ah, h] > g > [AH1, AG1]


The following usable rules [FROCOS05] were oriented:

a__g(X) → a__h(X)
a__cd
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__cc

(6) Obligation:

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

A__G(X) → A__H(X)

The TRS R consists of the following rules:

a__g(X) → a__h(X)
a__cd
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__cc

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) TRUE