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(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → 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(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → 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)) → B(b(a(c(x1))))
B(x1) → A(c(x1))
A(b(x1)) → A(c(x1))
B(x1) → C(x1)
A(b(x1)) → C(x1)
A(b(x1)) → B(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → 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)) → B(b(a(c(x1))))
B(x1) → A(c(x1))
A(b(x1)) → A(c(x1))
B(x1) → C(x1)
A(b(x1)) → C(x1)
A(b(x1)) → B(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → 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
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(a(c(x1))))
B(x1) → A(c(x1))
A(b(x1)) → A(c(x1))
A(b(x1)) → B(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(x1) → A(c(x1)) at position [0] we obtained the following new rules:
B(c(x0)) → A(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(a(c(x1))))
B(c(x0)) → A(x0)
A(b(x1)) → A(c(x1))
A(b(x1)) → B(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → A(c(x1)) at position [0] we obtained the following new rules:
A(b(c(x0))) → A(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(a(c(x1))))
B(c(x0)) → A(x0)
A(b(x1)) → B(a(c(x1)))
A(b(c(x0))) → A(x0)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → B(b(a(c(x1)))) at position [0] we obtained the following new rules:
A(b(c(x0))) → B(b(a(x0)))
A(b(y0)) → B(b(c(y0)))
A(b(y0)) → B(a(c(a(c(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(c(x0))) → B(b(a(x0)))
A(b(y0)) → B(a(c(a(c(y0)))))
A(b(y0)) → B(b(c(y0)))
B(c(x0)) → A(x0)
A(b(x1)) → B(a(c(x1)))
A(b(c(x0))) → A(x0)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → B(a(c(x1))) at position [0] we obtained the following new rules:
A(b(c(x0))) → B(a(x0))
A(b(y0)) → B(c(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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))) → B(b(a(x0)))
B(c(x0)) → A(x0)
A(b(y0)) → B(b(c(y0)))
A(b(y0)) → B(a(c(a(c(y0)))))
A(b(y0)) → B(c(y0))
A(b(c(x0))) → B(a(x0))
A(b(c(x0))) → A(x0)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → 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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → x1
A(b(c(x0))) → B(b(a(x0)))
B(c(x0)) → A(x0)
A(b(y0)) → B(b(c(y0)))
A(b(y0)) → B(a(c(a(c(y0)))))
A(b(y0)) → B(c(y0))
A(b(c(x0))) → B(a(x0))
A(b(c(x0))) → A(x0)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → x1
A(b(c(x0))) → B(b(a(x0)))
B(c(x0)) → A(x0)
A(b(y0)) → B(b(c(y0)))
A(b(y0)) → B(a(c(a(c(y0)))))
A(b(y0)) → B(c(y0))
A(b(c(x0))) → B(a(x0))
A(b(c(x0))) → A(x0)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
c(b(A(x))) → a(b(B(x)))
c(B(x)) → A(x)
b(A(x)) → c(b(B(x)))
b(A(x)) → c(a(c(a(B(x)))))
b(A(x)) → c(B(x))
c(b(A(x))) → a(B(x))
c(b(A(x))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
c(b(A(x))) → a(b(B(x)))
c(B(x)) → A(x)
b(A(x)) → c(b(B(x)))
b(A(x)) → c(a(c(a(B(x)))))
b(A(x)) → c(B(x))
c(b(A(x))) → a(B(x))
c(b(A(x))) → 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)) → C(a(B(x)))
C(b(A(x))) → B1(B(x))
B1(a(x)) → B1(x)
B1(A(x)) → B1(B(x))
B1(A(x)) → A1(c(a(B(x))))
B1(A(x)) → C(b(B(x)))
B1(a(x)) → C(a(b(b(x))))
B1(a(x)) → B1(b(x))
B1(A(x)) → C(B(x))
B1(x) → C(a(x))
B1(A(x)) → C(a(c(a(B(x)))))
B1(A(x)) → A1(B(x))
C(b(A(x))) → A1(B(x))
B1(x) → A1(x)
B1(a(x)) → A1(b(b(x)))
C(b(A(x))) → A1(b(B(x)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
c(b(A(x))) → a(b(B(x)))
c(B(x)) → A(x)
b(A(x)) → c(b(B(x)))
b(A(x)) → c(a(c(a(B(x)))))
b(A(x)) → c(B(x))
c(b(A(x))) → a(B(x))
c(b(A(x))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(A(x)) → C(a(B(x)))
C(b(A(x))) → B1(B(x))
B1(a(x)) → B1(x)
B1(A(x)) → B1(B(x))
B1(A(x)) → A1(c(a(B(x))))
B1(A(x)) → C(b(B(x)))
B1(a(x)) → C(a(b(b(x))))
B1(a(x)) → B1(b(x))
B1(A(x)) → C(B(x))
B1(x) → C(a(x))
B1(A(x)) → C(a(c(a(B(x)))))
B1(A(x)) → A1(B(x))
C(b(A(x))) → A1(B(x))
B1(x) → A1(x)
B1(a(x)) → A1(b(b(x)))
C(b(A(x))) → A1(b(B(x)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
c(b(A(x))) → a(b(B(x)))
c(B(x)) → A(x)
b(A(x)) → c(b(B(x)))
b(A(x)) → c(a(c(a(B(x)))))
b(A(x)) → c(B(x))
c(b(A(x))) → a(B(x))
c(b(A(x))) → 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 2 SCCs with 12 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(A(x))) → B1(B(x))
B1(x) → C(a(x))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
c(b(A(x))) → a(b(B(x)))
c(B(x)) → A(x)
b(A(x)) → c(b(B(x)))
b(A(x)) → c(a(c(a(B(x)))))
b(A(x)) → c(B(x))
c(b(A(x))) → a(B(x))
c(b(A(x))) → A(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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(x) → C(a(x))
C(b(A(x))) → B1(B(x))
The TRS R consists of the following rules:
a(x) → x
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
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(A(x))) → B1(B(x))
B1(x) → C(a(x))
The TRS R consists of the following rules:
a(x) → x
The set Q consists of the following terms:
a(x0)
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:
C(b(A(x))) → B1(B(x))
B1(x) → C(a(x))
The following rules are removed from R:
a(x) → x
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 1 + 2·x1
POL(B(x1)) = x1
POL(B1(x1)) = 2 + 2·x1
POL(C(x1)) = x1
POL(a(x1)) = 1 + x1
POL(b(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:
a(x0)
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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(x) → C(a(x))
C(b(A(x))) → B1(B(x))
The TRS R consists of the following rules:
a(x) → x
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
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(A(x))) → B1(B(x))
B1(x) → C(a(x))
The TRS R consists of the following rules:
a(x) → x
The set Q consists of the following terms:
a(x0)
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(x) → C(a(x))
C(b(A(x))) → B1(B(x))
The TRS R consists of the following rules:
a(x) → x
Q is empty.
We have to consider all (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(b(x))
B1(a(x)) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
c(b(A(x))) → a(b(B(x)))
c(B(x)) → A(x)
b(A(x)) → c(b(B(x)))
b(A(x)) → c(a(c(a(B(x)))))
b(A(x)) → c(B(x))
c(b(A(x))) → a(B(x))
c(b(A(x))) → 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) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
c(b(A(x))) → a(b(B(x)))
c(B(x)) → A(x)
b(A(x)) → c(b(B(x)))
b(A(x)) → c(a(c(a(B(x)))))
b(A(x)) → c(B(x))
c(b(A(x))) → a(B(x))
c(b(A(x))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → b(b(a(c(x))))
b(x) → a(c(x))
c(c(x)) → x
A(b(c(x))) → B(b(a(x)))
B(c(x)) → A(x)
A(b(x)) → B(b(c(x)))
A(b(x)) → B(a(c(a(c(x)))))
A(b(x)) → B(c(x))
A(b(c(x))) → B(a(x))
A(b(c(x))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ 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(b(a(c(x))))
b(x) → a(c(x))
c(c(x)) → x
A(b(c(x))) → B(b(a(x)))
B(c(x)) → A(x)
A(b(x)) → B(b(c(x)))
A(b(x)) → B(a(c(a(c(x)))))
A(b(x)) → B(c(x))
A(b(c(x))) → B(a(x))
A(b(c(x))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
c(b(A(x))) → a(b(B(x)))
c(B(x)) → A(x)
b(A(x)) → c(b(B(x)))
b(A(x)) → c(a(c(a(B(x)))))
b(A(x)) → c(B(x))
c(b(A(x))) → a(B(x))
c(b(A(x))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → b(b(a(c(x))))
b(x) → a(c(x))
c(c(x)) → x
A(b(c(x))) → B(b(a(x)))
B(c(x)) → A(x)
A(b(x)) → B(b(c(x)))
A(b(x)) → B(a(c(a(c(x)))))
A(b(x)) → B(c(x))
A(b(c(x))) → B(a(x))
A(b(c(x))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ 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(b(a(c(x))))
b(x) → a(c(x))
c(c(x)) → x
A(b(c(x))) → B(b(a(x)))
B(c(x)) → A(x)
A(b(x)) → B(b(c(x)))
A(b(x)) → B(a(c(a(c(x)))))
A(b(x)) → B(c(x))
A(b(c(x))) → B(a(x))
A(b(c(x))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → 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)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(b(a(c(x1))))
b(x1) → a(c(x1))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → 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)) → c(a(b(b(x))))
b(x) → c(a(x))
c(c(x)) → x
Q is empty.