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(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(c(x1))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(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(b(x1)) → A(c(x1))
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
A(c(x1)) → C(a(a(x1)))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → A(c(x1))
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
A(c(x1)) → C(a(a(x1)))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(C(x1)) = 2 + 2·x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → A(c(x1))
A(c(x1)) → C(a(a(x1)))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(x1)) → C(a(a(x1))) at position [0] we obtained the following new rules:
A(c(c(x0))) → C(a(b(c(a(a(x0))))))
A(c(b(x0))) → C(a(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(c(x0))) → C(a(b(c(a(a(x0))))))
C(b(x1)) → A(c(x1))
A(c(b(x0))) → C(a(x0))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(c(x1))
A(c(c(x0))) → C(a(b(c(a(a(x0))))))
C(b(x1)) → A(c(x1))
A(c(b(x0))) → C(a(x0))
C(b(x1)) → C(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(c(x1))
A(c(c(x0))) → C(a(b(c(a(a(x0))))))
C(b(x1)) → A(c(x1))
A(c(b(x0))) → C(a(x0))
C(b(x1)) → C(x1)
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(C(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ 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:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(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:
B(C(x)) → C1(A(x))
C1(a(x)) → B(x)
B(c(x)) → C1(a(x))
C1(c(A(x))) → B(a(C(x)))
C1(a(x)) → C1(b(x))
C1(c(A(x))) → C1(b(a(C(x))))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ 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(C(x)) → C1(A(x))
C1(a(x)) → B(x)
B(c(x)) → C1(a(x))
C1(c(A(x))) → B(a(C(x)))
C1(a(x)) → C1(b(x))
C1(c(A(x))) → C1(b(a(C(x))))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(x)) → B(x)
B(c(x)) → C1(a(x))
C1(a(x)) → C1(b(x))
C1(c(A(x))) → C1(b(a(C(x))))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B(c(x)) → C1(a(x))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(B(x1)) = 1 + x1
POL(C(x1)) = 1 + 2·x1
POL(C1(x1)) = 1 + x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(x)) → B(x)
C1(a(x)) → C1(b(x))
C1(c(A(x))) → C1(b(a(C(x))))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(x)) → C1(b(x))
C1(c(A(x))) → C1(b(a(C(x))))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(A(x))) → C1(b(a(C(x)))) at position [0] we obtained the following new rules:
C1(c(A(y0))) → C1(C(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ 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(a(x)) → C1(b(x))
C1(c(A(y0))) → C1(C(y0))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ 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(a(x)) → C1(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(x)) → C1(b(x)) at position [0] we obtained the following new rules:
C1(a(c(A(x0)))) → C1(a(C(x0)))
C1(a(c(x0))) → C1(c(a(x0)))
C1(a(C(x0))) → C1(C(x0))
C1(a(C(x0))) → C1(c(A(x0)))
C1(a(a(x0))) → C1(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ 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:
C1(a(c(x0))) → C1(c(a(x0)))
C1(a(C(x0))) → C1(C(x0))
C1(a(a(x0))) → C1(x0)
C1(a(c(A(x0)))) → C1(a(C(x0)))
C1(a(C(x0))) → C1(c(A(x0)))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ 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:
C1(a(c(x0))) → C1(c(a(x0)))
C1(a(a(x0))) → C1(x0)
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(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
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → x
a(c(x)) → b(c(a(a(x))))
c(b(x)) → a(c(x))
A(c(c(x))) → C(a(b(c(a(a(x))))))
C(b(x)) → A(c(x))
A(c(b(x))) → C(a(x))
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ 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(b(x)) → x
a(c(x)) → b(c(a(a(x))))
c(b(x)) → a(c(x))
A(c(c(x))) → C(a(b(c(a(a(x))))))
C(b(x)) → A(c(x))
A(c(b(x))) → C(a(x))
C(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
c(c(A(x))) → a(a(c(b(a(C(x))))))
b(C(x)) → c(A(x))
b(c(A(x))) → a(C(x))
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → x
a(c(x)) → b(c(a(a(x))))
c(b(x)) → a(c(x))
A(c(c(x))) → C(a(b(c(a(a(x))))))
C(b(x)) → A(c(x))
A(c(b(x))) → C(a(x))
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ 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(b(x)) → x
a(c(x)) → b(c(a(a(x))))
c(b(x)) → a(c(x))
A(c(c(x))) → C(a(b(c(a(a(x))))))
C(b(x)) → A(c(x))
A(c(b(x))) → C(a(x))
C(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(c(x1))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → 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:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(c(x1)) → b(c(a(a(x1))))
c(b(x1)) → a(c(x1))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → 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:
b(a(x)) → x
c(a(x)) → a(a(c(b(x))))
b(c(x)) → c(a(x))
Q is empty.