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(b(c(x1)))
c(c(a(x1))) → a(a(c(c(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(b(c(x1)))
c(c(a(x1))) → a(a(c(c(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(a(x1))) → A(c(c(x1)))
A(x1) → C(x1)
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
C(c(a(x1))) → A(a(c(c(x1))))
The TRS R consists of the following rules:
a(x1) → x1
a(x1) → b(b(c(x1)))
c(c(a(x1))) → a(a(c(c(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(c(a(x1))) → A(c(c(x1)))
A(x1) → C(x1)
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
C(c(a(x1))) → A(a(c(c(x1))))
The TRS R consists of the following rules:
a(x1) → x1
a(x1) → b(b(c(x1)))
c(c(a(x1))) → a(a(c(c(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(x1) → b(b(c(x1)))
c(c(a(x1))) → a(a(c(c(x1))))
C(c(a(x1))) → A(c(c(x1)))
A(x1) → C(x1)
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
C(c(a(x1))) → A(a(c(c(x1))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(x1) → b(b(c(x1)))
c(c(a(x1))) → a(a(c(c(x1))))
C(c(a(x1))) → A(c(c(x1)))
A(x1) → C(x1)
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
C(c(a(x1))) → A(a(c(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(a(A(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
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(a(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:
A1(c(c(x))) → A1(a(x))
A1(c(C(x))) → A2(x)
A1(c(C(x))) → A1(A(x))
A1(c(c(x))) → A1(x)
The TRS R consists of the following rules:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(a(A(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(c(c(x))) → A1(a(x))
A1(c(C(x))) → A2(x)
A1(c(C(x))) → A1(A(x))
A1(c(c(x))) → A1(x)
The TRS R consists of the following rules:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(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
↳ 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(c(c(x))) → A1(a(x))
A1(c(C(x))) → A1(A(x))
A1(c(c(x))) → A1(x)
The TRS R consists of the following rules:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(c(C(x))) → A1(A(x)) at position [0] we obtained the following new rules:
A1(c(C(x0))) → A1(C(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:
A1(c(c(x))) → A1(a(x))
A1(c(c(x))) → A1(x)
A1(c(C(x0))) → A1(C(x0))
The TRS R consists of the following rules:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(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
↳ 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:
A1(c(c(x))) → A1(a(x))
A1(c(c(x))) → A1(x)
The TRS R consists of the following rules:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(c(c(x))) → A1(a(x)) at position [0] we obtained the following new rules:
A1(c(c(c(C(x0))))) → A1(c(C(x0)))
A1(c(c(c(C(x0))))) → A1(c(c(a(A(x0)))))
A1(c(c(c(c(x0))))) → A1(c(c(a(a(x0)))))
A1(c(c(x0))) → A1(x0)
A1(c(c(c(C(x0))))) → A1(c(c(A(x0))))
A1(c(c(c(C(x0))))) → A1(C(x0))
A1(c(c(x0))) → A1(c(b(b(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(c(c(c(C(x0))))) → A1(c(c(a(A(x0)))))
A1(c(c(c(C(x0))))) → A1(c(C(x0)))
A1(c(c(c(c(x0))))) → A1(c(c(a(a(x0)))))
A1(c(c(x))) → A1(x)
A1(c(c(c(C(x0))))) → A1(c(c(A(x0))))
A1(c(c(c(C(x0))))) → A1(C(x0))
A1(c(c(x0))) → A1(c(b(b(x0))))
The TRS R consists of the following rules:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(c(c(c(C(x0))))) → A1(c(c(a(A(x0)))))
A1(c(c(c(c(x0))))) → A1(c(c(a(a(x0)))))
A1(c(c(x))) → A1(x)
A1(c(c(c(C(x0))))) → A1(c(c(A(x0))))
The TRS R consists of the following rules:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(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
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(a(A(x))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → b(b(c(x)))
c(c(a(x))) → a(a(c(c(x))))
C(c(a(x))) → A(c(c(x)))
A(x) → C(x)
C(c(a(x))) → C(c(x))
C(c(a(x))) → C(x)
C(c(a(x))) → A(a(c(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) → x
a(x) → b(b(c(x)))
c(c(a(x))) → a(a(c(c(x))))
C(c(a(x))) → A(c(c(x)))
A(x) → C(x)
C(c(a(x))) → C(c(x))
C(c(a(x))) → C(x)
C(c(a(x))) → A(a(c(c(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(a(A(x))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → b(b(c(x)))
c(c(a(x))) → a(a(c(c(x))))
C(c(a(x))) → A(c(c(x)))
A(x) → C(x)
C(c(a(x))) → C(c(x))
C(c(a(x))) → C(x)
C(c(a(x))) → A(a(c(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) → x
a(x) → b(b(c(x)))
c(c(a(x))) → a(a(c(c(x))))
C(c(a(x))) → A(c(c(x)))
A(x) → C(x)
C(c(a(x))) → C(c(x))
C(c(a(x))) → C(x)
C(c(a(x))) → A(a(c(c(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(x1) → b(b(c(x1)))
c(c(a(x1))) → a(a(c(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(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) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(x1) → b(b(c(x1)))
c(c(a(x1))) → a(a(c(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(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) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
Q is empty.