(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)
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(g(X), Y) → A__F(mark(X), f(g(X), Y))
A__F(g(X), Y) → MARK(X)
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → MARK(X)
The TRS R consists of the following rules:
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
A__F(g(X), Y) → MARK(X)
MARK(g(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__F(
x1,
x2) =
x1
g(
x1) =
g(
x1)
mark(
x1) =
x1
f(
x1,
x2) =
x1
MARK(
x1) =
x1
a__f(
x1,
x2) =
x1
Lexicographic Path Order [LPO].
Precedence:
trivial
The following usable rules [FROCOS05] were oriented:
mark(f(X1, X2)) → a__f(mark(X1), X2)
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)
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(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(
x1) =
x1
f(
x1,
x2) =
f(
x1,
x2)
Lexicographic Path Order [LPO].
Precedence:
trivial
The following usable rules [FROCOS05] were oriented:
none
(8) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)
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