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(a(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → c(a(x1))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → c(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:
C(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
C(b(x1)) → C(a(x1))
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → c(a(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
C(b(x1)) → C(a(x1))
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → c(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 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → c(a(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(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:
A(a(b(x1))) → A(x1)
Strictly oriented rules of the TRS R:
a(x1) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(a(b(x1))) → b(b(a(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(a(b(x1))) → b(b(a(a(x1))))
The set Q consists of the following terms:
a(a(b(x0)))
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(a(b(x1))) → A(a(x1))
Strictly oriented rules of the TRS R:
a(a(b(x1))) → b(b(a(a(x1))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(a(x1)) = 2·x1
POL(b(x1)) = 1 + x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:
a(a(b(x0)))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → C(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → c(a(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → C(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(x1)) → C(a(x1)) at position [0] we obtained the following new rules:
C(b(a(b(x0)))) → C(b(b(a(a(x0)))))
C(b(x0)) → C(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(a(b(x0)))) → C(b(b(a(a(x0)))))
C(b(x0)) → C(x0)
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(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
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
C(b(a(b(x0)))) → C(b(b(a(a(x0)))))
C(b(x0)) → C(x0)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
C(b(a(b(x0)))) → C(b(b(a(a(x0)))))
C(b(x0)) → C(x0)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(b(x))) → b(b(a(a(x))))
C(b(a(b(x)))) → C(b(b(a(a(x)))))
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(a(b(x))) → b(b(a(a(x))))
C(b(a(b(x)))) → C(b(b(a(a(x)))))
C(b(x)) → C(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(a(a(x))) → A(a(b(b(x))))
B(a(b(C(x)))) → A(b(b(C(x))))
B(a(a(x))) → A(b(b(x)))
B(a(b(C(x)))) → A(a(b(b(C(x)))))
B(a(a(x))) → B(x)
B(a(b(C(x)))) → B(b(C(x)))
B(a(a(x))) → B(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(x))) → A(a(b(b(x))))
B(a(b(C(x)))) → A(b(b(C(x))))
B(a(a(x))) → A(b(b(x)))
B(a(b(C(x)))) → A(a(b(b(C(x)))))
B(a(a(x))) → B(x)
B(a(b(C(x)))) → B(b(C(x)))
B(a(a(x))) → B(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(x))) → B(x)
B(a(b(C(x)))) → B(b(C(x)))
B(a(a(x))) → B(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(b(C(x)))) → B(b(C(x))) at position [0] we obtained the following new rules:
B(a(b(C(x0)))) → B(C(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(C(x0)))) → B(C(x0))
B(a(a(x))) → B(x)
B(a(a(x))) → B(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(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
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(x))) → B(x)
B(a(a(x))) → B(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(x))) → B(b(x)) at position [0] we obtained the following new rules:
B(a(a(a(a(x0))))) → B(a(a(b(b(x0)))))
B(a(a(C(x0)))) → B(C(x0))
B(a(a(a(b(C(x0)))))) → B(a(a(b(b(C(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(a(x0))))) → B(a(a(b(b(x0)))))
B(a(a(C(x0)))) → B(C(x0))
B(a(a(a(b(C(x0)))))) → B(a(a(b(b(C(x0))))))
B(a(a(x))) → B(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(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
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ UsableRulesProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(a(x0))))) → B(a(a(b(b(x0)))))
B(a(a(a(b(C(x0)))))) → B(a(a(b(b(C(x0))))))
B(a(a(x))) → B(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(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(a(x))) → a(a(b(b(x))))
b(a(b(C(x)))) → a(a(b(b(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(b(x))) → b(b(a(a(x))))
C(b(a(b(x)))) → C(b(b(a(a(x)))))
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ UsableRulesProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(a(b(x))) → b(b(a(a(x))))
C(b(a(b(x)))) → C(b(b(a(a(x)))))
C(b(x)) → C(x)
Q is empty.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → C(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
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(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → c(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(c(x)) → a(c(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(c(x)) → a(c(x))
Q is empty.