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(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(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(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(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:
B(a(x1)) → A(b(x1))
B(a(x1)) → A(c(a(b(x1))))
B(a(x1)) → C(a(b(x1)))
A(x1) → B(x1)
B(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(x1)) → A(b(x1))
B(a(x1)) → A(c(a(b(x1))))
B(a(x1)) → C(a(b(x1)))
A(x1) → B(x1)
B(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(x1)) → 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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(x1)) → A(b(x1))
B(a(x1)) → A(c(a(b(x1))))
A(x1) → B(x1)
B(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(x1)) → x1
B(a(x1)) → A(b(x1))
B(a(x1)) → A(c(a(b(x1))))
A(x1) → B(x1)
B(a(x1)) → B(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(x1)) → x1
B(a(x1)) → A(b(x1))
B(a(x1)) → A(c(a(b(x1))))
A(x1) → B(x1)
B(a(x1)) → B(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(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(B(x)) → A2(x)
A1(B(x)) → A1(c(A(x)))
A1(x) → B1(x)
A1(b(x)) → C(a(x))
A1(b(x)) → B1(a(c(a(x))))
A1(b(x)) → A1(c(a(x)))
A1(B(x)) → C(A(x))
A1(B(x)) → B1(a(c(A(x))))
A1(B(x)) → B1(A(x))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
A1(B(x)) → A2(x)
A1(B(x)) → A1(c(A(x)))
A1(x) → B1(x)
A1(b(x)) → C(a(x))
A1(b(x)) → B1(a(c(a(x))))
A1(b(x)) → A1(c(a(x)))
A1(B(x)) → C(A(x))
A1(B(x)) → B1(a(c(A(x))))
A1(B(x)) → B1(A(x))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(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
↳ 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(B(x)) → A1(c(A(x)))
A1(b(x)) → A1(c(a(x)))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(B(x)) → A1(c(A(x))) at position [0] we obtained the following new rules:
A1(B(x0)) → A1(c(B(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ 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:
A1(B(x0)) → A1(c(B(x0)))
A1(b(x)) → A1(c(a(x)))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(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
↳ 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:
A1(b(x)) → A1(c(a(x)))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(x)) → A1(c(a(x))) at position [0] we obtained the following new rules:
A1(b(B(x0))) → A1(c(b(A(x0))))
A1(b(x0)) → A1(c(b(x0)))
A1(b(B(x0))) → A1(c(B(x0)))
A1(b(b(x0))) → A1(c(b(a(c(a(x0))))))
A1(b(B(x0))) → A1(c(b(a(c(A(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(B(x0))) → A1(c(b(A(x0))))
A1(b(x0)) → A1(c(b(x0)))
A1(b(b(x0))) → A1(c(b(a(c(a(x0))))))
A1(b(B(x0))) → A1(c(B(x0)))
A1(b(B(x0))) → A1(c(b(a(c(A(x0))))))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(b(B(x0))) → A1(c(b(A(x0))))
A1(b(x0)) → A1(c(b(x0)))
A1(b(b(x0))) → A1(c(b(a(c(a(x0))))))
A1(b(B(x0))) → A1(c(b(a(c(A(x0))))))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(B(x0))) → A1(c(b(A(x0)))) at position [0] we obtained the following new rules:
A1(b(B(x0))) → A1(c(b(B(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(b(B(x0))) → A1(c(b(B(x0))))
A1(b(x0)) → A1(c(b(x0)))
A1(b(b(x0))) → A1(c(b(a(c(a(x0))))))
A1(b(x)) → A1(x)
A1(b(B(x0))) → A1(c(b(a(c(A(x0))))))
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(b(x0)) → A1(c(b(x0)))
A1(b(b(x0))) → A1(c(b(a(c(a(x0))))))
A1(b(B(x0))) → A1(c(b(a(c(A(x0))))))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(x0)) → A1(c(b(x0))) at position [0] we obtained the following new rules:
A1(b(a(x0))) → A1(c(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(x0))) → A1(c(x0))
A1(b(b(x0))) → A1(c(b(a(c(a(x0))))))
A1(b(x)) → A1(x)
A1(b(B(x0))) → A1(c(b(a(c(A(x0))))))
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(x0))) → A1(c(x0)) at position [0] we obtained the following new rules:
A1(b(a(c(x0)))) → A1(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(b(x0))) → A1(c(b(a(c(a(x0))))))
A1(b(a(c(x0)))) → A1(x0)
A1(b(B(x0))) → A1(c(b(a(c(A(x0))))))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(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) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(b(x)) → x
b(a(x)) → a(c(a(b(x))))
c(c(x)) → x
B(a(x)) → A(b(x))
B(a(x)) → A(c(a(b(x))))
A(x) → B(x)
B(a(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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) → b(x)
a(b(x)) → x
b(a(x)) → a(c(a(b(x))))
c(c(x)) → x
B(a(x)) → A(b(x))
B(a(x)) → A(c(a(b(x))))
A(x) → B(x)
B(a(x)) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(B(x)) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(b(x)) → x
b(a(x)) → a(c(a(b(x))))
c(c(x)) → x
B(a(x)) → A(b(x))
B(a(x)) → A(c(a(b(x))))
A(x) → B(x)
B(a(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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) → b(x)
a(b(x)) → x
b(a(x)) → a(c(a(b(x))))
c(c(x)) → x
B(a(x)) → A(b(x))
B(a(x)) → A(c(a(b(x))))
A(x) → B(x)
B(a(x)) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(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) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(x1)) → x1
b(a(x1)) → a(c(a(b(x1))))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(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) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
Q is empty.