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:
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(a(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(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:
A(d(x1)) → A(x1)
C(a(x1)) → C(x1)
A(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))
B(d(x1)) → A(b(x1))
A(a(x1)) → B(a(x1))
C(b(x1)) → A(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
B(a(a(x1))) → A(b(c(x1)))
B(d(x1)) → B(x1)
C(b(x1)) → B(a(x1))
C(a(x1)) → A(c(x1))
A(a(b(x1))) → B(a(x1))
The TRS R consists of the following rules:
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(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:
A(d(x1)) → A(x1)
C(a(x1)) → C(x1)
A(a(b(x1))) → A(x1)
A(a(x1)) → A(b(a(x1)))
B(d(x1)) → A(b(x1))
A(a(x1)) → B(a(x1))
C(b(x1)) → A(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
B(a(a(x1))) → A(b(c(x1)))
B(d(x1)) → B(x1)
C(b(x1)) → B(a(x1))
C(a(x1)) → A(c(x1))
A(a(b(x1))) → B(a(x1))
The TRS R consists of the following rules:
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(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(d(x1)) → A(x1)
C(a(x1)) → C(x1)
A(a(b(x1))) → A(x1)
A(a(x1)) → B(a(x1))
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
B(d(x1)) → B(x1)
A(a(b(x1))) → B(a(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2 + x1
POL(B(x1)) = x1
POL(C(x1)) = 2 + x1
POL(a(x1)) = 2 + x1
POL(b(x1)) = x1
POL(c(x1)) = 2 + x1
POL(d(x1)) = 2 + x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → A(x1)
A(a(x1)) → A(b(a(x1)))
B(a(a(x1))) → A(b(c(x1)))
B(d(x1)) → A(b(x1))
C(b(x1)) → B(a(x1))
C(a(x1)) → A(c(x1))
The TRS R consists of the following rules:
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(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 1 SCC with 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(x1)) → A(b(a(x1)))
The TRS R consists of the following rules:
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(x1)) → A(b(a(x1))) at position [0] we obtained the following new rules:
A(a(a(b(x0)))) → A(b(d(b(a(x0)))))
A(a(a(x0))) → A(a(b(c(x0))))
A(a(d(x0))) → A(b(d(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(x0)))) → A(b(d(b(a(x0)))))
A(a(d(x0))) → A(b(d(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(a(x0))) → A(a(b(c(x0))))
The TRS R consists of the following rules:
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(a(x1)))
A(a(a(b(x0)))) → A(b(d(b(a(x0)))))
A(a(d(x0))) → A(b(d(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(a(x0))) → A(a(b(c(x0))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(a(x1)))
A(a(a(b(x0)))) → A(b(d(b(a(x0)))))
A(a(d(x0))) → A(b(d(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(a(x0))) → A(a(b(c(x0))))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(a(A(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(a(A(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(a(A(x))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
a(a(b(x))) → d(b(a(x)))
a(d(x)) → d(a(x))
b(d(x)) → a(b(x))
a(a(x)) → a(b(a(x)))
A(a(a(b(x)))) → A(b(d(b(a(x)))))
A(a(d(x))) → A(b(d(a(x))))
A(a(a(x))) → A(b(a(b(a(x)))))
A(a(a(x))) → A(a(b(c(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
a(a(b(x))) → d(b(a(x)))
a(d(x)) → d(a(x))
b(d(x)) → a(b(x))
a(a(x)) → a(b(a(x)))
A(a(a(b(x)))) → A(b(d(b(a(x)))))
A(a(d(x))) → A(b(d(a(x))))
A(a(a(x))) → A(b(a(b(a(x)))))
A(a(a(x))) → A(a(b(c(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(a(A(x))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
a(a(b(x))) → d(b(a(x)))
a(d(x)) → d(a(x))
b(d(x)) → a(b(x))
a(a(x)) → a(b(a(x)))
A(a(a(b(x)))) → A(b(d(b(a(x)))))
A(a(d(x))) → A(b(d(a(x))))
A(a(a(x))) → A(b(a(b(a(x)))))
A(a(a(x))) → A(a(b(c(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
a(a(b(x))) → d(b(a(x)))
a(d(x)) → d(a(x))
b(d(x)) → a(b(x))
a(a(x)) → a(b(a(x)))
A(a(a(b(x)))) → A(b(d(b(a(x)))))
A(a(d(x))) → A(b(d(a(x))))
A(a(a(x))) → A(b(a(b(a(x)))))
A(a(a(x))) → A(a(b(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:
D(a(A(x))) → B(A(x))
D(a(A(x))) → A1(d(b(A(x))))
A1(a(A(x))) → A1(b(a(b(A(x)))))
A1(a(x)) → A1(b(a(x)))
A1(a(A(x))) → A1(b(A(x)))
B(a(a(A(x)))) → D(b(A(x)))
A1(a(A(x))) → B(a(b(A(x))))
B(a(a(x))) → B(d(x))
B(c(x)) → A1(b(x))
A1(c(x)) → A1(x)
B(a(a(A(x)))) → A1(b(d(b(A(x)))))
D(b(x)) → A1(x)
A1(a(b(x))) → B(a(x))
B(c(x)) → B(x)
A1(a(x)) → B(a(x))
B(a(a(x))) → A1(b(d(x)))
B(a(a(A(x)))) → B(A(x))
A1(a(A(x))) → B(a(A(x)))
B(a(a(x))) → D(x)
D(a(A(x))) → D(b(A(x)))
B(a(a(A(x)))) → B(d(b(A(x))))
A1(a(b(x))) → A1(x)
D(a(x)) → D(x)
D(a(x)) → A1(d(x))
A1(a(A(x))) → B(A(x))
D(b(x)) → B(a(x))
The TRS R consists of the following rules:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(a(A(x))) → B(A(x))
D(a(A(x))) → A1(d(b(A(x))))
A1(a(A(x))) → A1(b(a(b(A(x)))))
A1(a(x)) → A1(b(a(x)))
A1(a(A(x))) → A1(b(A(x)))
B(a(a(A(x)))) → D(b(A(x)))
A1(a(A(x))) → B(a(b(A(x))))
B(a(a(x))) → B(d(x))
B(c(x)) → A1(b(x))
A1(c(x)) → A1(x)
B(a(a(A(x)))) → A1(b(d(b(A(x)))))
D(b(x)) → A1(x)
A1(a(b(x))) → B(a(x))
B(c(x)) → B(x)
A1(a(x)) → B(a(x))
B(a(a(x))) → A1(b(d(x)))
B(a(a(A(x)))) → B(A(x))
A1(a(A(x))) → B(a(A(x)))
B(a(a(x))) → D(x)
D(a(A(x))) → D(b(A(x)))
B(a(a(A(x)))) → B(d(b(A(x))))
A1(a(b(x))) → A1(x)
D(a(x)) → D(x)
D(a(x)) → A1(d(x))
A1(a(A(x))) → B(A(x))
D(b(x)) → B(a(x))
The TRS R consists of the following rules:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(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 7 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(x))) → A1(b(d(x)))
B(a(a(x))) → D(x)
D(a(A(x))) → D(b(A(x)))
D(a(A(x))) → A1(d(b(A(x))))
B(a(a(A(x)))) → B(d(b(A(x))))
A1(a(x)) → A1(b(a(x)))
A1(a(b(x))) → A1(x)
D(a(x)) → D(x)
B(a(a(A(x)))) → D(b(A(x)))
D(a(x)) → A1(d(x))
B(a(a(x))) → B(d(x))
B(c(x)) → A1(b(x))
A1(c(x)) → A1(x)
D(b(x)) → B(a(x))
B(a(a(A(x)))) → A1(b(d(b(A(x)))))
D(b(x)) → A1(x)
A1(a(b(x))) → B(a(x))
B(c(x)) → B(x)
A1(a(x)) → B(a(x))
The TRS R consists of the following rules:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(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:
B(a(a(x))) → A1(b(d(x)))
B(a(a(x))) → D(x)
D(a(A(x))) → D(b(A(x)))
D(a(A(x))) → A1(d(b(A(x))))
B(a(a(A(x)))) → B(d(b(A(x))))
A1(a(b(x))) → A1(x)
D(a(x)) → D(x)
B(a(a(A(x)))) → D(b(A(x)))
D(a(x)) → A1(d(x))
B(a(a(x))) → B(d(x))
B(c(x)) → A1(b(x))
A1(c(x)) → A1(x)
B(a(a(A(x)))) → A1(b(d(b(A(x)))))
D(b(x)) → A1(x)
A1(a(b(x))) → B(a(x))
B(c(x)) → B(x)
A1(a(x)) → B(a(x))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(A1(x1)) = 2·x1
POL(B(x1)) = x1
POL(D(x1)) = 2 + 2·x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = x1
POL(c(x1)) = 2 + 2·x1
POL(d(x1)) = 2 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(b(x)) → B(a(x))
A1(a(x)) → A1(b(a(x)))
The TRS R consists of the following rules:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → A1(b(a(x)))
The TRS R consists of the following rules:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(x)) → A1(b(a(x))) at position [0] we obtained the following new rules:
A1(a(a(A(x0)))) → A1(a(b(d(b(A(x0))))))
A1(a(c(x0))) → A1(b(c(a(x0))))
A1(a(a(A(x0)))) → A1(b(a(b(a(b(A(x0)))))))
A1(a(a(A(x0)))) → A1(b(c(b(a(A(x0))))))
A1(a(a(x0))) → A1(a(b(d(x0))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(b(x0)))) → A1(b(c(b(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(A(x0)))) → A1(a(b(d(b(A(x0))))))
A1(a(a(A(x0)))) → A1(b(a(b(a(b(A(x0)))))))
A1(a(c(x0))) → A1(b(c(a(x0))))
A1(a(a(A(x0)))) → A1(b(c(b(a(A(x0))))))
A1(a(a(x0))) → A1(a(b(d(x0))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(b(x0)))) → A1(b(c(b(a(x0)))))
The TRS R consists of the following rules:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(c(x0))) → A1(b(c(a(x0))))
A1(a(a(x0))) → A1(a(b(d(x0))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(b(x0)))) → A1(b(c(b(a(x0)))))
The TRS R consists of the following rules:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
b(a(a(A(x)))) → a(b(d(b(A(x)))))
d(a(A(x))) → a(d(b(A(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(A(x))) → c(b(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
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(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(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
a(a(b(x1))) → d(b(a(x1)))
a(d(x1)) → d(a(x1))
b(d(x1)) → a(b(x1))
a(a(x1)) → a(b(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(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(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(x)))
Q is empty.