(0) Obligation:

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

a__f(f(a)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
mark(g(X)) → g(mark(X))
a__f(X) → f(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:

MARK(f(X)) → A__F(mark(X))
MARK(f(X)) → MARK(X)
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(a)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
mark(g(X)) → g(mark(X))
a__f(X) → f(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 1 less node.

(4) Obligation:

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

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

The TRS R consists of the following rules:

a__f(f(a)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
mark(g(X)) → g(mark(X))
a__f(X) → f(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(g(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)  =  MARK(x1)
g(x1)  =  g(x1)
f(x1)  =  f(x1)
a__f(x1)  =  a__f(x1)
a  =  a
c(x1)  =  c
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
MARK1 > c
mark1 > [g1, af1, a] > f1 > c

Status:
MARK1: [1]
g1: [1]
f1: [1]
af1: [1]
a: []
c: []
mark1: [1]


The following usable rules [FROCOS05] were oriented:

a__f(f(a)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
mark(g(X)) → g(mark(X))
a__f(X) → f(X)

(6) Obligation:

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

a__f(f(a)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
mark(g(X)) → g(mark(X))
a__f(X) → f(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