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(b(c(x1)))
c(a(b(x1))) → a(a(c(x1)))
c(b(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) → b(b(c(x1)))
c(a(b(x1))) → a(a(c(x1)))
c(b(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:
C(a(b(x1))) → A(c(x1))
A(x1) → C(x1)
C(a(b(x1))) → A(a(c(x1)))
C(a(b(x1))) → C(x1)
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
c(a(b(x1))) → a(a(c(x1)))
c(b(x1)) → 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(a(b(x1))) → A(c(x1))
A(x1) → C(x1)
C(a(b(x1))) → A(a(c(x1)))
C(a(b(x1))) → C(x1)
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
c(a(b(x1))) → a(a(c(x1)))
c(b(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
c(a(b(x1))) → a(a(c(x1)))
c(b(x1)) → x1
C(a(b(x1))) → A(c(x1))
A(x1) → C(x1)
C(a(b(x1))) → A(a(c(x1)))
C(a(b(x1))) → C(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(c(x1)))
c(a(b(x1))) → a(a(c(x1)))
c(b(x1)) → x1
C(a(b(x1))) → A(c(x1))
A(x1) → C(x1)
C(a(b(x1))) → A(a(c(x1)))
C(a(b(x1))) → C(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
b(a(C(x))) → c(A(x))
A(x) → C(x)
b(a(C(x))) → c(a(A(x)))
b(a(C(x))) → 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) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
b(a(C(x))) → c(A(x))
A(x) → C(x)
b(a(C(x))) → c(a(A(x)))
b(a(C(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:
A1(x) → B(x)
B(a(c(x))) → A1(x)
B(a(C(x))) → A2(x)
B(a(C(x))) → A1(A(x))
A1(x) → B(b(x))
B(a(c(x))) → A1(a(x))
The TRS R consists of the following rules:
a(x) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
b(a(C(x))) → c(A(x))
A(x) → C(x)
b(a(C(x))) → c(a(A(x)))
b(a(C(x))) → 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:
A1(x) → B(x)
B(a(c(x))) → A1(x)
B(a(C(x))) → A2(x)
B(a(C(x))) → A1(A(x))
A1(x) → B(b(x))
B(a(c(x))) → A1(a(x))
The TRS R consists of the following rules:
a(x) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
b(a(C(x))) → c(A(x))
A(x) → C(x)
b(a(C(x))) → c(a(A(x)))
b(a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(a(c(x))) → A1(x)
B(a(C(x))) → A1(A(x))
A1(x) → B(b(x))
B(a(c(x))) → A1(a(x))
The TRS R consists of the following rules:
a(x) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
b(a(C(x))) → c(A(x))
A(x) → C(x)
b(a(C(x))) → c(a(A(x)))
b(a(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) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
b(a(C(x))) → c(A(x))
A(x) → C(x)
b(a(C(x))) → c(a(A(x)))
b(a(C(x))) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(b(c(x)))
c(a(b(x))) → a(a(c(x)))
c(b(x)) → x
C(a(b(x))) → A(c(x))
A(x) → C(x)
C(a(b(x))) → A(a(c(x)))
C(a(b(x))) → C(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) → b(b(c(x)))
c(a(b(x))) → a(a(c(x)))
c(b(x)) → x
C(a(b(x))) → A(c(x))
A(x) → C(x)
C(a(b(x))) → A(a(c(x)))
C(a(b(x))) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
b(a(C(x))) → c(A(x))
A(x) → C(x)
b(a(C(x))) → c(a(A(x)))
b(a(C(x))) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(b(c(x)))
c(a(b(x))) → a(a(c(x)))
c(b(x)) → x
C(a(b(x))) → A(c(x))
A(x) → C(x)
C(a(b(x))) → A(a(c(x)))
C(a(b(x))) → C(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) → b(b(c(x)))
c(a(b(x))) → a(a(c(x)))
c(b(x)) → x
C(a(b(x))) → A(c(x))
A(x) → C(x)
C(a(b(x))) → A(a(c(x)))
C(a(b(x))) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(c(x1)))
c(a(b(x1))) → a(a(c(x1)))
c(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(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) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(c(x1)))
c(a(b(x1))) → a(a(c(x1)))
c(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(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) → c(b(b(x)))
b(a(c(x))) → c(a(a(x)))
b(c(x)) → x
Q is empty.