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(a(b(x1)))) → c(b(c(a(x1))))
c(x1) → b(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) → x1
a(b(a(b(x1)))) → c(b(c(a(x1))))
c(x1) → 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:
C(x1) → A(x1)
A(b(a(b(x1)))) → C(b(c(a(x1))))
A(b(a(b(x1)))) → A(x1)
A(b(a(b(x1)))) → C(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(b(a(b(x1)))) → c(b(c(a(x1))))
c(x1) → b(a(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(x1) → A(x1)
A(b(a(b(x1)))) → C(b(c(a(x1))))
A(b(a(b(x1)))) → A(x1)
A(b(a(b(x1)))) → C(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(b(a(b(x1)))) → c(b(c(a(x1))))
c(x1) → 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
↳ 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(a(b(x1)))) → c(b(c(a(x1))))
c(x1) → b(a(x1))
C(x1) → A(x1)
A(b(a(b(x1)))) → C(b(c(a(x1))))
A(b(a(b(x1)))) → A(x1)
A(b(a(b(x1)))) → C(a(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(a(b(x1)))) → c(b(c(a(x1))))
c(x1) → b(a(x1))
C(x1) → A(x1)
A(b(a(b(x1)))) → C(b(c(a(x1))))
A(b(a(b(x1)))) → A(x1)
A(b(a(b(x1)))) → C(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(C(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(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(b(A(x)))) → B(C(x))
B(a(b(a(x)))) → B(c(x))
C1(x) → B(x)
B(a(b(A(x)))) → C2(x)
B(a(b(A(x)))) → C1(b(C(x)))
B(a(b(a(x)))) → A1(c(b(c(x))))
B(a(b(a(x)))) → C1(b(c(x)))
B(a(b(A(x)))) → A1(c(b(C(x))))
C1(x) → A1(b(x))
B(a(b(A(x)))) → A1(C(x))
B(a(b(a(x)))) → C1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(C(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(a(b(A(x)))) → B(C(x))
B(a(b(a(x)))) → B(c(x))
C1(x) → B(x)
B(a(b(A(x)))) → C2(x)
B(a(b(A(x)))) → C1(b(C(x)))
B(a(b(a(x)))) → A1(c(b(c(x))))
B(a(b(a(x)))) → C1(b(c(x)))
B(a(b(A(x)))) → A1(c(b(C(x))))
C1(x) → A1(b(x))
B(a(b(A(x)))) → A1(C(x))
B(a(b(a(x)))) → C1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(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 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(A(x)))) → B(C(x))
B(a(b(a(x)))) → B(c(x))
C1(x) → B(x)
B(a(b(A(x)))) → C1(b(C(x)))
B(a(b(a(x)))) → C1(b(c(x)))
B(a(b(a(x)))) → C1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(C(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(b(A(x)))) → B(C(x)) at position [0] we obtained the following new rules:
B(a(b(A(x0)))) → B(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(x)))) → B(c(x))
C1(x) → B(x)
B(a(b(A(x)))) → C1(b(C(x)))
B(a(b(a(x)))) → C1(b(c(x)))
B(a(b(A(x0)))) → B(A(x0))
B(a(b(a(x)))) → C1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(x)))) → B(c(x))
C1(x) → B(x)
B(a(b(A(x)))) → C1(b(C(x)))
B(a(b(a(x)))) → C1(b(c(x)))
B(a(b(a(x)))) → C1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(C(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(b(a(x)))) → B(c(x)) at position [0] we obtained the following new rules:
B(a(b(a(x0)))) → B(a(b(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(x0)))) → B(a(b(x0)))
C1(x) → B(x)
B(a(b(A(x)))) → C1(b(C(x)))
B(a(b(a(x)))) → C1(b(c(x)))
B(a(b(a(x)))) → C1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(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(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(C(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(a(b(x)))) → c(b(c(a(x))))
c(x) → b(a(x))
C(x) → A(x)
A(b(a(b(x)))) → C(b(c(a(x))))
A(b(a(b(x)))) → A(x)
A(b(a(b(x)))) → C(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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(a(b(x)))) → c(b(c(a(x))))
c(x) → b(a(x))
C(x) → A(x)
A(b(a(b(x)))) → C(b(c(a(x))))
A(b(a(b(x)))) → A(x)
A(b(a(b(x)))) → C(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
C(x) → A(x)
b(a(b(A(x)))) → a(c(b(C(x))))
b(a(b(A(x)))) → A(x)
b(a(b(A(x)))) → a(C(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(a(b(x)))) → c(b(c(a(x))))
c(x) → b(a(x))
C(x) → A(x)
A(b(a(b(x)))) → C(b(c(a(x))))
A(b(a(b(x)))) → A(x)
A(b(a(b(x)))) → C(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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(a(b(x)))) → c(b(c(a(x))))
c(x) → b(a(x))
C(x) → A(x)
A(b(a(b(x)))) → C(b(c(a(x))))
A(b(a(b(x)))) → A(x)
A(b(a(b(x)))) → C(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(a(b(x1)))) → c(b(c(a(x1))))
c(x1) → b(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(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) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(a(b(x1)))) → c(b(c(a(x1))))
c(x1) → b(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(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) → x
b(a(b(a(x)))) → a(c(b(c(x))))
c(x) → a(b(x))
Q is empty.