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(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(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(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(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(c(x1)) → A(x1)
C(c(x1)) → A(c(a(x1)))
C(c(x1)) → C(a(x1))
A(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(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(c(x1)) → A(x1)
C(c(x1)) → A(c(a(x1)))
C(c(x1)) → C(a(x1))
A(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(x1)) → A(c(a(x1))) at position [0] we obtained the following new rules:
C(c(x0)) → A(c(b(x0)))
C(c(b(x0))) → A(c(b(c(x0))))
C(c(x0)) → A(c(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(x1)) → A(x1)
C(c(b(x0))) → A(c(b(c(x0))))
C(c(x0)) → A(c(x0))
C(c(x0)) → A(c(b(x0)))
C(c(x1)) → C(a(x1))
A(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(x1)) → A(x1)
C(c(x0)) → A(c(x0))
C(c(x1)) → C(a(x1))
A(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(x1)) → C(a(x1)) at position [0] we obtained the following new rules:
C(c(x0)) → C(b(x0))
C(c(x0)) → C(x0)
C(c(b(x0))) → C(b(c(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(x0)) → C(x0)
C(c(x0)) → C(b(x0))
C(c(x1)) → A(x1)
C(c(x0)) → A(c(x0))
C(c(b(x0))) → C(b(c(x0)))
A(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(x0)) → C(x0)
C(c(x1)) → A(x1)
C(c(x0)) → A(c(x0))
A(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(a(x1)))
C(c(x0)) → C(x0)
C(c(x1)) → A(x1)
C(c(x0)) → A(c(x0))
A(b(x1)) → C(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(a(x1)))
C(c(x0)) → C(x0)
C(c(x1)) → A(x1)
C(c(x0)) → A(c(x0))
A(b(x1)) → C(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(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:
C1(c(x)) → A1(c(a(x)))
A1(x) → B(x)
C1(c(x)) → A1(x)
C1(C(x)) → C1(A(x))
B(a(x)) → B(x)
B(a(x)) → C1(b(x))
C1(c(x)) → C1(a(x))
The TRS R consists of the following rules:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C1(c(x)) → A1(c(a(x)))
A1(x) → B(x)
C1(c(x)) → A1(x)
C1(C(x)) → C1(A(x))
B(a(x)) → B(x)
B(a(x)) → C1(b(x))
C1(c(x)) → C1(a(x))
The TRS R consists of the following rules:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(x) → B(x)
C1(c(x)) → A1(c(a(x)))
C1(c(x)) → A1(x)
B(a(x)) → B(x)
B(a(x)) → C1(b(x))
C1(c(x)) → C1(a(x))
The TRS R consists of the following rules:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(x)) → C1(b(x)) at position [0] we obtained the following new rules:
B(a(A(x0))) → C1(C(x0))
B(a(a(x0))) → C1(c(b(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C1(c(x)) → A1(c(a(x)))
A1(x) → B(x)
B(a(a(x0))) → C1(c(b(x0)))
C1(c(x)) → A1(x)
B(a(A(x0))) → C1(C(x0))
B(a(x)) → B(x)
C1(c(x)) → C1(a(x))
The TRS R consists of the following rules:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C1(c(x)) → A1(c(a(x)))
A1(x) → B(x)
B(a(a(x0))) → C1(c(b(x0)))
C1(c(x)) → A1(x)
B(a(x)) → B(x)
C1(c(x)) → C1(a(x))
The TRS R consists of the following rules:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(x)) → C1(a(x)) at position [0] we obtained the following new rules:
C1(c(x0)) → C1(x0)
C1(c(x0)) → C1(b(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(x) → B(x)
C1(c(x)) → A1(c(a(x)))
B(a(a(x0))) → C1(c(b(x0)))
C1(c(x0)) → C1(x0)
C1(c(x)) → A1(x)
B(a(x)) → B(x)
C1(c(x0)) → C1(b(x0))
The TRS R consists of the following rules:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(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) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → b(x)
a(b(x)) → b(c(x))
c(c(x)) → a(c(a(x)))
C(c(x)) → C(x)
C(c(x)) → A(x)
C(c(x)) → A(c(x))
A(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(x) → b(x)
a(b(x)) → b(c(x))
c(c(x)) → a(c(a(x)))
C(c(x)) → C(x)
C(c(x)) → A(x)
C(c(x)) → A(c(x))
A(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(A(x))
b(A(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → b(x)
a(b(x)) → b(c(x))
c(c(x)) → a(c(a(x)))
C(c(x)) → C(x)
C(c(x)) → A(x)
C(c(x)) → A(c(x))
A(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(x) → b(x)
a(b(x)) → b(c(x))
c(c(x)) → a(c(a(x)))
C(c(x)) → C(x)
C(c(x)) → A(x)
C(c(x)) → A(c(x))
A(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(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) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(x1) → b(x1)
a(b(x1)) → b(c(x1))
c(c(x1)) → a(c(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(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) → x
a(x) → b(x)
b(a(x)) → c(b(x))
c(c(x)) → a(c(a(x)))
Q is empty.