(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__g(X) → a__h(X)
a__c → d
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__c → c
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__c → d
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__c → c
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__c → d
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__c → c
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) =
x1
d =
d
A__G(
x1) =
x1
c =
c
a__g(
x1) =
a__g(
x1)
a__h(
x1) =
a__h(
x1)
a__c =
a__c
mark(
x1) =
mark(
x1)
g(
x1) =
x1
h(
x1) =
x1
Recursive Path Order [RPO].
Precedence:
mark1 > ac > d > c > ah1
mark1 > ac > d > ag1 > ah1
The following usable rules [FROCOS05] were oriented:
a__g(X) → a__h(X)
a__c → d
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__c → c
(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__c → d
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__c → c
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