Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, b, X) → a__f(mark(X), X, mark(X))
a__c → a
a__c → b
mark(f(X1, X2, X3)) → a__f(mark(X1), X2, mark(X3))
mark(c) → a__c
mark(a) → a
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(a, b, X) → a__f(mark(X), X, mark(X))
a__c → a
a__c → b
mark(f(X1, X2, X3)) → a__f(mark(X1), X2, mark(X3))
mark(c) → a__c
mark(a) → a
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X1, X2, X3)) → MARK(X3)
A__F(a, b, X) → MARK(X)
MARK(c) → A__C
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))
A__F(a, b, X) → A__F(mark(X), X, mark(X))
The TRS R consists of the following rules:
a__f(a, b, X) → a__f(mark(X), X, mark(X))
a__c → a
a__c → b
mark(f(X1, X2, X3)) → a__f(mark(X1), X2, mark(X3))
mark(c) → a__c
mark(a) → a
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X1, X2, X3)) → MARK(X3)
A__F(a, b, X) → MARK(X)
MARK(c) → A__C
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))
A__F(a, b, X) → A__F(mark(X), X, mark(X))
The TRS R consists of the following rules:
a__f(a, b, X) → a__f(mark(X), X, mark(X))
a__c → a
a__c → b
mark(f(X1, X2, X3)) → a__f(mark(X1), X2, mark(X3))
mark(c) → a__c
mark(a) → a
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X1, X2, X3)) → MARK(X3)
A__F(a, b, X) → MARK(X)
MARK(f(X1, X2, X3)) → MARK(X1)
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))
The TRS R consists of the following rules:
a__f(a, b, X) → a__f(mark(X), X, mark(X))
a__c → a
a__c → b
mark(f(X1, X2, X3)) → a__f(mark(X1), X2, mark(X3))
mark(c) → a__c
mark(a) → a
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c
Q is empty.
We have to consider all minimal (P,Q,R)-chains.