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(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → 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) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → 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(b(x1))) → A(x1)
A(c(x1)) → A(x1)
A(c(x1)) → C(a(x1))
A(c(x1)) → C(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x1))) → A(x1)
A(c(x1)) → A(x1)
A(c(x1)) → C(a(x1))
A(c(x1)) → C(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(x1)) → C(c(a(x1))) at position [0] we obtained the following new rules:
A(c(x0)) → C(c(b(x0)))
A(c(c(x0))) → C(c(b(c(c(a(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(x0)) → C(c(b(x0)))
C(b(b(x1))) → A(x1)
A(c(x1)) → A(x1)
A(c(c(x0))) → C(c(b(c(c(a(x0))))))
A(c(x1)) → C(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(x1)) → C(a(x1)) at position [0] we obtained the following new rules:
A(c(c(x0))) → C(b(c(c(a(x0)))))
A(c(x0)) → C(b(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x1))) → A(x1)
A(c(x0)) → C(c(b(x0)))
A(c(x0)) → C(b(x0))
A(c(x1)) → A(x1)
A(c(c(x0))) → C(c(b(c(c(a(x0))))))
A(c(c(x0))) → C(b(c(c(a(x0)))))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A(c(x1)) → A(x1) we obtained the following new rules:
A(c(c(y_0))) → A(c(y_0))
A(c(c(c(y_0)))) → A(c(c(y_0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(c(y_0))) → A(c(y_0))
A(c(x0)) → C(c(b(x0)))
C(b(b(x1))) → A(x1)
A(c(c(c(y_0)))) → A(c(c(y_0)))
A(c(x0)) → C(b(x0))
A(c(c(x0))) → C(c(b(c(c(a(x0))))))
A(c(c(x0))) → C(b(c(c(a(x0)))))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule C(b(b(x1))) → A(x1) we obtained the following new rules:
C(b(b(c(y_0)))) → A(c(y_0))
C(b(b(c(c(y_0))))) → A(c(c(y_0)))
C(b(b(c(c(c(y_0)))))) → A(c(c(c(y_0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(c(y_0))) → A(c(y_0))
A(c(x0)) → C(c(b(x0)))
A(c(c(c(y_0)))) → A(c(c(y_0)))
C(b(b(c(y_0)))) → A(c(y_0))
A(c(x0)) → C(b(x0))
C(b(b(c(c(y_0))))) → A(c(c(y_0)))
C(b(b(c(c(c(y_0)))))) → A(c(c(c(y_0))))
A(c(c(x0))) → C(c(b(c(c(a(x0))))))
A(c(c(x0))) → C(b(c(c(a(x0)))))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → 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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
A(c(c(y_0))) → A(c(y_0))
A(c(x0)) → C(c(b(x0)))
A(c(c(c(y_0)))) → A(c(c(y_0)))
C(b(b(c(y_0)))) → A(c(y_0))
A(c(x0)) → C(b(x0))
C(b(b(c(c(y_0))))) → A(c(c(y_0)))
C(b(b(c(c(c(y_0)))))) → A(c(c(c(y_0))))
A(c(c(x0))) → C(c(b(c(c(a(x0))))))
A(c(c(x0))) → C(b(c(c(a(x0)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
A(c(c(y_0))) → A(c(y_0))
A(c(x0)) → C(c(b(x0)))
A(c(c(c(y_0)))) → A(c(c(y_0)))
C(b(b(c(y_0)))) → A(c(y_0))
A(c(x0)) → C(b(x0))
C(b(b(c(c(y_0))))) → A(c(c(y_0)))
C(b(b(c(c(c(y_0)))))) → A(c(c(c(y_0))))
A(c(c(x0))) → C(c(b(c(c(a(x0))))))
A(c(c(x0))) → C(b(c(c(a(x0)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(c(x)) → b(c(c(a(x))))
c(b(b(x))) → a(x)
A(c(c(x))) → A(c(x))
A(c(x)) → C(c(b(x)))
A(c(c(c(x)))) → A(c(c(x)))
C(b(b(c(x)))) → A(c(x))
A(c(x)) → C(b(x))
C(b(b(c(c(x))))) → A(c(c(x)))
C(b(b(c(c(c(x)))))) → A(c(c(c(x))))
A(c(c(x))) → C(c(b(c(c(a(x))))))
A(c(c(x))) → C(b(c(c(a(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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(c(x)) → b(c(c(a(x))))
c(b(b(x))) → a(x)
A(c(c(x))) → A(c(x))
A(c(x)) → C(c(b(x)))
A(c(c(c(x)))) → A(c(c(x)))
C(b(b(c(x)))) → A(c(x))
A(c(x)) → C(b(x))
C(b(b(c(c(x))))) → A(c(c(x)))
C(b(b(c(c(c(x)))))) → A(c(c(c(x))))
A(c(c(x))) → C(c(b(c(c(a(x))))))
A(c(c(x))) → C(b(c(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:
C1(c(A(x))) → B(c(C(x)))
C1(c(A(x))) → B(C(x))
C1(a(x)) → B(x)
C1(A(x)) → B(C(x))
C1(c(A(x))) → C1(b(c(C(x))))
C1(c(A(x))) → A1(c(c(b(C(x)))))
C1(c(A(x))) → A1(c(c(b(c(C(x))))))
C1(c(c(b(b(C(x)))))) → C1(A(x))
C1(c(b(b(C(x))))) → C1(A(x))
C1(c(A(x))) → C1(c(b(c(C(x)))))
C1(a(x)) → C1(b(x))
C1(a(x)) → A1(c(c(b(x))))
C1(c(A(x))) → C1(C(x))
C1(b(b(C(x)))) → C1(A(x))
C1(c(A(x))) → C1(b(C(x)))
A1(x) → B(x)
C1(a(x)) → C1(c(b(x)))
C1(A(x)) → B(c(C(x)))
B(b(c(x))) → A1(x)
C1(A(x)) → C1(C(x))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(A(x))) → C1(c(b(C(x))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(c(b(b(C(x))))) → C1(c(A(x)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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:
C1(c(A(x))) → B(c(C(x)))
C1(c(A(x))) → B(C(x))
C1(a(x)) → B(x)
C1(A(x)) → B(C(x))
C1(c(A(x))) → C1(b(c(C(x))))
C1(c(A(x))) → A1(c(c(b(C(x)))))
C1(c(A(x))) → A1(c(c(b(c(C(x))))))
C1(c(c(b(b(C(x)))))) → C1(A(x))
C1(c(b(b(C(x))))) → C1(A(x))
C1(c(A(x))) → C1(c(b(c(C(x)))))
C1(a(x)) → C1(b(x))
C1(a(x)) → A1(c(c(b(x))))
C1(c(A(x))) → C1(C(x))
C1(b(b(C(x)))) → C1(A(x))
C1(c(A(x))) → C1(b(C(x)))
A1(x) → B(x)
C1(a(x)) → C1(c(b(x)))
C1(A(x)) → B(c(C(x)))
B(b(c(x))) → A1(x)
C1(A(x)) → C1(C(x))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(A(x))) → C1(c(b(C(x))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(c(b(b(C(x))))) → C1(c(A(x)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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 2 SCCs with 17 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(b(c(x))) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(b(c(x))) → A1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
A1(x) → B(x)
B(b(c(x))) → A1(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A1(x1)) = 2 + x1
POL(B(x1)) = 1 + x1
POL(b(x1)) = 2 + x1
POL(c(x1)) = 2 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(b(c(x))) → A1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(a(x)) → C1(b(x))
C1(c(b(b(C(x))))) → C1(c(A(x)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(b(b(C(x))))) → C1(c(A(x))) at position [0] we obtained the following new rules:
C1(c(b(b(C(x0))))) → C1(b(C(x0)))
C1(c(b(b(C(x0))))) → C1(b(c(C(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(b(b(C(x0))))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(c(b(b(C(x0))))) → C1(b(C(x0)))
C1(a(x)) → C1(b(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(a(x)) → C1(b(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x)))))) → C1(c(A(x))) at position [0] we obtained the following new rules:
C1(c(c(b(b(C(x0)))))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(b(C(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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:
C1(c(c(b(b(C(x0)))))) → C1(b(c(C(x0))))
C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(a(x)) → C1(b(x))
C1(c(c(b(b(C(x0)))))) → C1(b(C(x0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(a(x)) → C1(b(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(x)) → C1(b(x)) at position [0] we obtained the following new rules:
C1(a(b(c(x0)))) → C1(a(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(a(b(c(x0)))) → C1(a(x0))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(x)) → C1(c(b(x))) at position [0] we obtained the following new rules:
C1(a(b(C(x0)))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(c(a(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(a(b(C(x0)))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x)))))) → C1(c(c(A(x)))) at position [0] we obtained the following new rules:
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(c(c(b(b(C(x0)))))) → C1(c(b(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(c(b(c(C(x0)))))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(a(b(C(x0)))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))
C1(c(c(b(b(C(x0)))))) → C1(c(b(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(c(b(c(C(x0)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(a(b(C(x0)))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(b(C(x0)))) → C1(c(A(x0))) at position [0] we obtained the following new rules:
C1(a(b(C(x0)))) → C1(b(c(C(x0))))
C1(a(b(C(x0)))) → C1(b(C(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(C(x0)))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))
C1(a(b(C(x0)))) → C1(b(C(x0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x0)))))) → C1(c(A(x0))) at position [0] we obtained the following new rules:
C1(c(c(b(b(C(x0)))))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(b(C(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C1(c(c(b(b(C(x0)))))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(c(x0)))) → C1(a(x0))
C1(c(c(b(b(C(x0)))))) → C1(b(C(x0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(c(x0)))) → C1(a(x0))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0)))))) at position [0] we obtained the following new rules:
C1(c(c(b(b(C(y0)))))) → C1(b(c(c(b(C(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(c(x0)))) → C1(a(x0))
C1(c(c(b(b(C(y0)))))) → C1(b(c(c(b(C(y0))))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(c(x0)))) → C1(a(x0))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0))))))) at position [0] we obtained the following new rules:
C1(c(c(b(b(C(y0)))))) → C1(b(c(c(b(c(C(y0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(y0)))))) → C1(b(c(c(b(c(C(y0)))))))
C1(a(b(c(x0)))) → C1(a(x0))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(a(b(c(x0)))) → C1(a(x0))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(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) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
c(c(A(x))) → c(A(x))
c(A(x)) → b(c(C(x)))
c(c(c(A(x)))) → c(c(A(x)))
c(b(b(C(x)))) → c(A(x))
c(A(x)) → b(C(x))
c(c(b(b(C(x))))) → c(c(A(x)))
c(c(c(b(b(C(x)))))) → c(c(c(A(x))))
c(c(A(x))) → a(c(c(b(c(C(x))))))
c(c(A(x))) → a(c(c(b(C(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(c(x)) → b(c(c(a(x))))
c(b(b(x))) → a(x)
A(c(c(x))) → A(c(x))
A(c(x)) → C(c(b(x)))
A(c(c(c(x)))) → A(c(c(x)))
C(b(b(c(x)))) → A(c(x))
A(c(x)) → C(b(x))
C(b(b(c(c(x))))) → A(c(c(x)))
C(b(b(c(c(c(x)))))) → A(c(c(c(x))))
A(c(c(x))) → C(c(b(c(c(a(x))))))
A(c(c(x))) → C(b(c(c(a(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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(c(x)) → b(c(c(a(x))))
c(b(b(x))) → a(x)
A(c(c(x))) → A(c(x))
A(c(x)) → C(c(b(x)))
A(c(c(c(x)))) → A(c(c(x)))
C(b(b(c(x)))) → A(c(x))
A(c(x)) → C(b(x))
C(b(b(c(c(x))))) → A(c(c(x)))
C(b(b(c(c(c(x)))))) → A(c(c(c(x))))
A(c(c(x))) → C(c(b(c(c(a(x))))))
A(c(c(x))) → C(b(c(c(a(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → 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(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(c(x1)) → b(c(c(a(x1))))
c(b(b(x1))) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → 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(x) → b(x)
c(a(x)) → a(c(c(b(x))))
b(b(c(x))) → a(x)
Q is empty.