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:
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(x1))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(x1))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(x1))))
The set Q is empty.
We have obtained the following QTRS:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(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:
11(0(0(1(x1)))) → 01(0(x1))
01(0(0(0(x1)))) → 11(1(x1))
01(0(0(0(x1)))) → 11(1(1(x1)))
11(0(0(1(x1)))) → 01(0(0(x1)))
11(0(0(1(x1)))) → 01(0(0(0(x1))))
01(0(0(0(x1)))) → 11(x1)
01(0(0(0(x1)))) → 01(1(1(1(x1))))
11(0(0(1(x1)))) → 01(x1)
The TRS R consists of the following rules:
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
11(0(0(1(x1)))) → 01(0(x1))
01(0(0(0(x1)))) → 11(1(x1))
01(0(0(0(x1)))) → 11(1(1(x1)))
11(0(0(1(x1)))) → 01(0(0(x1)))
11(0(0(1(x1)))) → 01(0(0(0(x1))))
01(0(0(0(x1)))) → 11(x1)
01(0(0(0(x1)))) → 01(1(1(1(x1))))
11(0(0(1(x1)))) → 01(x1)
The TRS R consists of the following rules:
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(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:
11(0(0(1(x1)))) → 01(0(x1))
01(0(0(0(x1)))) → 11(1(x1))
01(0(0(0(x1)))) → 11(1(1(x1)))
11(0(0(1(x1)))) → 01(0(0(x1)))
11(0(0(1(x1)))) → 01(0(0(0(x1))))
01(0(0(0(x1)))) → 11(x1)
11(0(0(1(x1)))) → 01(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(0(x1)) = 2 + 2·x1
POL(01(x1)) = 1 + 2·x1
POL(1(x1)) = 2 + 2·x1
POL(11(x1)) = 2 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
01(0(0(0(x1)))) → 01(1(1(1(x1))))
The TRS R consists of the following rules:
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 01(0(0(0(x1)))) → 01(1(1(1(x1)))) at position [0] we obtained the following new rules:
01(0(0(0(0(0(1(x0))))))) → 01(1(1(0(0(0(0(x0)))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
01(0(0(0(0(0(1(x0))))))) → 01(1(1(0(0(0(0(x0)))))))
The TRS R consists of the following rules:
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(x1))))
01(0(0(0(0(0(1(x0))))))) → 01(1(1(0(0(0(0(x0)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(x1))))
01(0(0(0(0(0(1(x0))))))) → 01(1(1(0(0(0(0(x0)))))))
The set Q is empty.
We have obtained the following QTRS:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(x)))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(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:
02(0(0(0(x)))) → 11(0(x))
11(0(0(1(x)))) → 02(0(0(x)))
02(0(0(0(x)))) → 11(1(1(0(x))))
11(0(0(0(0(0(01(x))))))) → 11(1(01(x)))
11(0(0(1(x)))) → 02(0(x))
11(0(0(0(0(0(01(x))))))) → 02(0(0(0(1(1(01(x)))))))
11(0(0(1(x)))) → 02(0(0(0(x))))
02(0(0(0(x)))) → 11(1(0(x)))
11(0(0(1(x)))) → 02(x)
11(0(0(0(0(0(01(x))))))) → 02(1(1(01(x))))
11(0(0(0(0(0(01(x))))))) → 02(0(1(1(01(x)))))
11(0(0(0(0(0(01(x))))))) → 02(0(0(1(1(01(x))))))
11(0(0(0(0(0(01(x))))))) → 11(01(x))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
02(0(0(0(x)))) → 11(0(x))
11(0(0(1(x)))) → 02(0(0(x)))
02(0(0(0(x)))) → 11(1(1(0(x))))
11(0(0(0(0(0(01(x))))))) → 11(1(01(x)))
11(0(0(1(x)))) → 02(0(x))
11(0(0(0(0(0(01(x))))))) → 02(0(0(0(1(1(01(x)))))))
11(0(0(1(x)))) → 02(0(0(0(x))))
02(0(0(0(x)))) → 11(1(0(x)))
11(0(0(1(x)))) → 02(x)
11(0(0(0(0(0(01(x))))))) → 02(1(1(01(x))))
11(0(0(0(0(0(01(x))))))) → 02(0(1(1(01(x)))))
11(0(0(0(0(0(01(x))))))) → 02(0(0(1(1(01(x))))))
11(0(0(0(0(0(01(x))))))) → 11(01(x))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(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 5 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
11(0(0(1(x)))) → 02(x)
02(0(0(0(x)))) → 11(0(x))
11(0(0(1(x)))) → 02(0(0(x)))
02(0(0(0(x)))) → 11(1(1(0(x))))
11(0(0(1(x)))) → 02(0(x))
11(0(0(0(0(0(01(x))))))) → 02(0(0(0(1(1(01(x)))))))
11(0(0(1(x)))) → 02(0(0(0(x))))
02(0(0(0(x)))) → 11(1(0(x)))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(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:
11(0(0(1(x)))) → 02(x)
02(0(0(0(x)))) → 11(0(x))
11(0(0(1(x)))) → 02(0(0(x)))
11(0(0(1(x)))) → 02(0(x))
02(0(0(0(x)))) → 11(1(0(x)))
Used ordering: POLO with Polynomial interpretation [25]:
POL(0(x1)) = 2 + 2·x1
POL(01(x1)) = x1
POL(02(x1)) = 2·x1
POL(1(x1)) = 2 + 2·x1
POL(11(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
02(0(0(0(x)))) → 11(1(1(0(x))))
11(0(0(0(0(0(01(x))))))) → 02(0(0(0(1(1(01(x)))))))
11(0(0(1(x)))) → 02(0(0(0(x))))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 02(0(0(0(x)))) → 11(1(1(0(x)))) at position [0] we obtained the following new rules:
02(0(0(0(0(0(0(x0))))))) → 11(1(1(1(1(1(0(x0)))))))
02(0(0(0(0(1(x0)))))) → 11(1(0(0(0(0(x0))))))
02(0(0(0(0(0(0(0(01(x0))))))))) → 11(1(0(0(0(0(1(1(01(x0)))))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
02(0(0(0(0(1(x0)))))) → 11(1(0(0(0(0(x0))))))
02(0(0(0(0(0(0(x0))))))) → 11(1(1(1(1(1(0(x0)))))))
11(0(0(0(0(0(01(x))))))) → 02(0(0(0(1(1(01(x)))))))
02(0(0(0(0(0(0(0(01(x0))))))))) → 11(1(0(0(0(0(1(1(01(x0)))))))))
11(0(0(1(x)))) → 02(0(0(0(x))))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
02(0(0(0(0(1(x0)))))) → 11(1(0(0(0(0(x0))))))
02(0(0(0(0(0(0(x0))))))) → 11(1(1(1(1(1(0(x0)))))))
02(0(0(0(0(0(0(0(01(x0))))))))) → 11(1(0(0(0(0(1(1(01(x0)))))))))
11(0(0(1(x)))) → 02(0(0(0(x))))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 02(0(0(0(0(0(0(0(01(x0))))))))) → 11(1(0(0(0(0(1(1(01(x0))))))))) at position [0] we obtained the following new rules:
02(0(0(0(0(0(0(0(01(y0))))))))) → 11(1(1(1(1(0(1(1(01(y0)))))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
02(0(0(0(0(0(0(x0))))))) → 11(1(1(1(1(1(0(x0)))))))
02(0(0(0(0(1(x0)))))) → 11(1(0(0(0(0(x0))))))
02(0(0(0(0(0(0(0(01(y0))))))))) → 11(1(1(1(1(0(1(1(01(y0)))))))))
11(0(0(1(x)))) → 02(0(0(0(x))))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
02(0(0(0(0(1(x0)))))) → 11(1(0(0(0(0(x0))))))
02(0(0(0(0(0(0(x0))))))) → 11(1(1(1(1(1(0(x0)))))))
11(0(0(1(x)))) → 02(0(0(0(x))))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(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
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(x)))))))
The set Q is empty.
We have obtained the following QTRS:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
01(0(0(0(0(0(1(x))))))) → 01(1(1(0(0(0(0(x)))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
01(0(0(0(0(0(1(x))))))) → 01(1(1(0(0(0(0(x)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(0(0(0(1(1(01(x)))))))
The set Q is empty.
We have obtained the following QTRS:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
01(0(0(0(0(0(1(x))))))) → 01(1(1(0(0(0(0(x)))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
01(0(0(0(0(0(1(x))))))) → 01(1(1(0(0(0(0(x)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(0(0(x1))))
The set Q is empty.
We have obtained the following QTRS:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(0(0(0(x))))
Q is empty.