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(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
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:
B(c(c(x1))) → A(x1)
A(b(x1)) → B(c(a(x1)))
B(c(c(x1))) → B(a(x1))
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(c(x1))) → A(x1)
A(b(x1)) → B(c(a(x1)))
B(c(c(x1))) → B(a(x1))
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
A(b(x1)) → B(c(a(x1)))
A(b(x1)) → A(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(B(x1)) = 2·x1
POL(a(x1)) = x1
POL(b(x1)) = 1 + 2·x1
POL(c(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(c(x1))) → A(x1)
B(c(c(x1))) → B(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
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
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(c(x1))) → B(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(c(x1))) → B(a(x1)) at position [0] we obtained the following new rules:
B(c(c(b(x0)))) → B(b(c(a(x0))))
B(c(c(x0))) → B(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(c(x0))) → B(x0)
B(c(c(b(x0)))) → B(b(c(a(x0))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
B(c(c(x0))) → B(x0)
B(c(c(b(x0)))) → B(b(c(a(x0))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
B(c(c(x0))) → B(x0)
B(c(c(b(x0)))) → B(b(c(a(x0))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → b(c(a(x)))
b(c(c(x))) → c(b(a(x)))
B(c(c(x))) → B(x)
B(c(c(b(x)))) → B(b(c(a(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(b(x)) → b(c(a(x)))
b(c(c(x))) → c(b(a(x)))
B(c(c(x))) → B(x)
B(c(c(b(x)))) → B(b(c(a(x))))
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:
B1(a(x)) → B1(x)
C(c(b(x))) → B1(c(x))
C(c(b(x))) → A(b(c(x)))
B1(a(x)) → C(b(x))
B1(c(c(B(x)))) → B1(B(x))
B1(c(c(B(x)))) → A(c(b(B(x))))
C(c(b(x))) → C(x)
B1(c(c(B(x)))) → C(b(B(x)))
B1(a(x)) → A(c(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
C(c(b(x))) → B1(c(x))
C(c(b(x))) → A(b(c(x)))
B1(a(x)) → C(b(x))
B1(c(c(B(x)))) → B1(B(x))
B1(c(c(B(x)))) → A(c(b(B(x))))
C(c(b(x))) → C(x)
B1(c(c(B(x)))) → C(b(B(x)))
B1(a(x)) → A(c(b(x)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
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 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
C(c(b(x))) → B1(c(x))
B1(a(x)) → C(b(x))
C(c(b(x))) → C(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
C(c(b(x))) → C(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(B(x1)) = 2·x1
POL(B1(x1)) = 2 + 2·x1
POL(C(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = 2 + 2·x1
POL(c(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(a(x)) → C(b(x))
C(c(b(x))) → B1(c(x))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(b(x))) → B1(c(x)) at position [0] we obtained the following new rules:
C(c(b(c(B(x0))))) → B1(B(x0))
C(c(b(c(b(x0))))) → B1(a(b(c(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
C(c(b(c(b(x0))))) → B1(a(b(c(x0))))
C(c(b(c(B(x0))))) → B1(B(x0))
B1(a(x)) → C(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
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
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
C(c(b(c(b(x0))))) → B1(a(b(c(x0))))
B1(a(x)) → C(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(x)) → C(b(x)) at position [0] we obtained the following new rules:
B1(a(a(x0))) → C(a(c(b(x0))))
B1(a(c(c(B(x0))))) → C(a(c(b(B(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(c(c(B(x0))))) → C(a(c(b(B(x0)))))
B1(a(x)) → B1(x)
C(c(b(c(b(x0))))) → B1(a(b(c(x0))))
B1(a(a(x0))) → C(a(c(b(x0))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(c(c(B(x0))))) → C(a(c(b(B(x0))))) at position [0] we obtained the following new rules:
B1(a(c(c(B(y0))))) → C(c(b(B(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(a(c(c(B(y0))))) → C(c(b(B(y0))))
C(c(b(c(b(x0))))) → B1(a(b(c(x0))))
B1(a(a(x0))) → C(a(c(b(x0))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
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
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
C(c(b(c(b(x0))))) → B1(a(b(c(x0))))
B1(a(a(x0))) → C(a(c(b(x0))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
c(c(B(x))) → B(x)
b(c(c(B(x)))) → a(c(b(B(x))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → b(c(a(x)))
b(c(c(x))) → c(b(a(x)))
B(c(c(x))) → B(x)
B(c(c(b(x)))) → B(b(c(a(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(b(x)) → b(c(a(x)))
b(c(c(x))) → c(b(a(x)))
B(c(c(x))) → B(x)
B(c(c(b(x)))) → B(b(c(a(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
b(c(c(x1))) → c(b(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(c(b(x))) → a(b(c(x)))
Q is empty.