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(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(a(x1))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(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(x1)) → A(x1)
A(a(x1)) → B(b(x1))
B(c(x1)) → A(a(x1))
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(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(x1)) → A(x1)
A(a(x1)) → B(b(x1))
B(c(x1)) → A(a(x1))
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(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:
B(c(x1)) → A(x1)
A(a(x1)) → B(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(B(x1)) = 2·x1
POL(a(x1)) = 1 + 2·x1
POL(b(x1)) = 1 + 2·x1
POL(c(x1)) = 1 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(x1)) → B(b(x1))
B(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(a(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(x1)) → B(b(x1)) at position [0] we obtained the following new rules:
A(a(c(x0))) → B(a(a(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(c(x0))) → B(a(a(x0)))
B(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(a(x1))
A(a(c(x0))) → B(a(a(x0)))
B(c(x1)) → A(a(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(a(x1))
A(a(c(x0))) → B(a(a(x0)))
B(c(x1)) → A(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ 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(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(b(x))
a(b(x)) → c(c(x))
b(c(x)) → a(a(x))
A(a(c(x))) → B(a(a(x)))
B(c(x)) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ 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(a(x)) → b(b(x))
a(b(x)) → c(c(x))
b(c(x)) → a(a(x))
A(a(c(x))) → B(a(a(x)))
B(c(x)) → A(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:
A1(a(x)) → B1(x)
C(a(A(x))) → A1(a(B(x)))
C(b(x)) → A1(x)
C(b(x)) → A1(a(x))
B1(a(x)) → C(x)
C(B(x)) → A1(A(x))
B1(a(x)) → C(c(x))
C(a(A(x))) → A1(B(x))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ 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:
A1(a(x)) → B1(x)
C(a(A(x))) → A1(a(B(x)))
C(b(x)) → A1(x)
C(b(x)) → A1(a(x))
B1(a(x)) → C(x)
C(B(x)) → A1(A(x))
B1(a(x)) → C(c(x))
C(a(A(x))) → A1(B(x))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ 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:
A1(a(x)) → B1(x)
C(a(A(x))) → A1(a(B(x)))
C(b(x)) → A1(x)
C(b(x)) → A1(a(x))
B1(a(x)) → C(x)
B1(a(x)) → C(c(x))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(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:
A1(a(x)) → B1(x)
C(b(x)) → A1(x)
B1(a(x)) → C(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(A1(x1)) = 2 + 2·x1
POL(B(x1)) = 2·x1
POL(B1(x1)) = 2 + 2·x1
POL(C(x1)) = 2 + 2·x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = 2 + 2·x1
POL(c(x1)) = 2 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ 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:
C(a(A(x))) → A1(a(B(x)))
C(b(x)) → A1(a(x))
B1(a(x)) → C(c(x))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(x)) → B1(b(x)) at position [0] we obtained the following new rules:
A1(a(a(x0))) → B1(c(c(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ 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:
C(a(A(x))) → A1(a(B(x)))
C(b(x)) → A1(a(x))
B1(a(x)) → C(c(x))
A1(a(a(x0))) → B1(c(c(x0)))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(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
↳ 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:
C(b(x)) → A1(a(x))
B1(a(x)) → C(c(x))
A1(a(a(x0))) → B1(c(c(x0)))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(x)) → C(c(x)) at position [0] we obtained the following new rules:
B1(a(b(x0))) → C(a(a(x0)))
B1(a(a(A(x0)))) → C(a(a(B(x0))))
B1(a(B(x0))) → C(a(A(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → A1(a(x))
B1(a(b(x0))) → C(a(a(x0)))
B1(a(a(A(x0)))) → C(a(a(B(x0))))
B1(a(B(x0))) → C(a(A(x0)))
A1(a(a(x0))) → B1(c(c(x0)))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → A1(a(x))
B1(a(b(x0))) → C(a(a(x0)))
B1(a(a(A(x0)))) → C(a(a(B(x0))))
A1(a(a(x0))) → B1(c(c(x0)))
The TRS R consists of the following rules:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(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(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
c(a(A(x))) → a(a(B(x)))
c(B(x)) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(b(x))
a(b(x)) → c(c(x))
b(c(x)) → a(a(x))
A(a(c(x))) → B(a(a(x)))
B(c(x)) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ 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(a(x)) → b(b(x))
a(b(x)) → c(c(x))
b(c(x)) → a(a(x))
A(a(c(x))) → B(a(a(x)))
B(c(x)) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(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(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(b(x1))
a(b(x1)) → c(c(x1))
b(c(x1)) → a(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(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(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(x))
Q is empty.