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