(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.
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A__F(x1, x2)) = x1
POL(MARK(x1)) = x1
POL(a__f(x1, x2)) = 1 + x1
POL(f(x1, x2)) = 1 + x1
POL(g(x1)) = x1
POL(mark(x1)) = x1
The following usable rules [FROCOS05] were oriented:
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
(4) 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(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.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
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.
(8) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(g(X)) → MARK(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- MARK(g(X)) → MARK(X)
The graph contains the following edges 1 > 1
(11) TRUE
(12) 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))
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.
(13) 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))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
| POL(A__F(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
| POL(f(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
| POL(a__f(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)
a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
(14) 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.
(15) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(16) TRUE