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:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
The set Q is empty.
We have obtained the following QTRS:
2(R(x)) → R(2(x))
3(R(x)) → R(3(x))
1(R(x)) → 3(L(x))
L(3(x)) → 3(L(x))
L(2(x)) → 2(L(x))
L(0(x)) → R(2(x))
b(R(x)) → b(1(c(x)))
c(3(x)) → 1(c(x))
1(c(2(x))) → 1(R(0(c(x))))
0(c(2(x))) → 0(0(c(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
2(R(x)) → R(2(x))
3(R(x)) → R(3(x))
1(R(x)) → 3(L(x))
L(3(x)) → 3(L(x))
L(2(x)) → 2(L(x))
L(0(x)) → R(2(x))
b(R(x)) → b(1(c(x)))
c(3(x)) → 1(c(x))
1(c(2(x))) → 1(R(0(c(x))))
0(c(2(x))) → 0(0(c(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
The set Q is empty.
We have obtained the following QTRS:
2(R(x)) → R(2(x))
3(R(x)) → R(3(x))
1(R(x)) → 3(L(x))
L(3(x)) → 3(L(x))
L(2(x)) → 2(L(x))
L(0(x)) → R(2(x))
b(R(x)) → b(1(c(x)))
c(3(x)) → 1(c(x))
1(c(2(x))) → 1(R(0(c(x))))
0(c(2(x))) → 0(0(c(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
2(R(x)) → R(2(x))
3(R(x)) → R(3(x))
1(R(x)) → 3(L(x))
L(3(x)) → 3(L(x))
L(2(x)) → 2(L(x))
L(0(x)) → R(2(x))
b(R(x)) → b(1(c(x)))
c(3(x)) → 1(c(x))
1(c(2(x))) → 1(R(0(c(x))))
0(c(2(x))) → 0(0(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:
21(c(1(x1))) → R1(1(x1))
21(L(x1)) → 21(x1)
21(c(0(x1))) → 01(0(x1))
21(c(1(x1))) → 01(R(1(x1)))
01(L(x1)) → 21(R(x1))
R1(1(x1)) → 31(x1)
R1(3(x1)) → 31(R(x1))
31(L(x1)) → 31(x1)
R1(3(x1)) → R1(x1)
R1(2(x1)) → 21(R(x1))
R1(2(x1)) → R1(x1)
01(L(x1)) → R1(x1)
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
21(c(1(x1))) → R1(1(x1))
21(L(x1)) → 21(x1)
21(c(0(x1))) → 01(0(x1))
21(c(1(x1))) → 01(R(1(x1)))
01(L(x1)) → 21(R(x1))
R1(1(x1)) → 31(x1)
R1(3(x1)) → 31(R(x1))
31(L(x1)) → 31(x1)
R1(3(x1)) → R1(x1)
R1(2(x1)) → 21(R(x1))
R1(2(x1)) → R1(x1)
01(L(x1)) → R1(x1)
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
31(L(x1)) → 31(x1)
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
31(L(x1)) → 31(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
31(L(x1)) → 31(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(31(x1)) = 2·x1
POL(L(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
31(L(x1)) → 31(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
21(L(x1)) → 21(x1)
21(c(0(x1))) → 01(0(x1))
01(L(x1)) → 21(R(x1))
21(c(1(x1))) → 01(R(1(x1)))
R1(3(x1)) → R1(x1)
R1(2(x1)) → 21(R(x1))
R1(2(x1)) → R1(x1)
01(L(x1)) → R1(x1)
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(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:
R1(2(x1)) → 21(R(x1))
R1(2(x1)) → R1(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(0(x1)) = 1 + x1
POL(01(x1)) = 1 + x1
POL(1(x1)) = x1
POL(2(x1)) = 1 + x1
POL(21(x1)) = 1 + x1
POL(3(x1)) = x1
POL(L(x1)) = x1
POL(R(x1)) = x1
POL(R1(x1)) = 1 + x1
POL(b(x1)) = x1
POL(c(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
21(L(x1)) → 21(x1)
21(c(1(x1))) → 01(R(1(x1)))
01(L(x1)) → 21(R(x1))
21(c(0(x1))) → 01(0(x1))
R1(3(x1)) → R1(x1)
01(L(x1)) → R1(x1)
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
R1(3(x1)) → R1(x1)
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
R1(3(x1)) → R1(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
R1(3(x1)) → R1(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(3(x1)) = 2·x1
POL(R1(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
21(L(x1)) → 21(x1)
21(c(0(x1))) → 01(0(x1))
01(L(x1)) → 21(R(x1))
21(c(1(x1))) → 01(R(1(x1)))
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 21(c(1(x1))) → 01(R(1(x1))) at position [0] we obtained the following new rules:
21(c(1(x0))) → 01(L(3(x0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
21(L(x1)) → 21(x1)
01(L(x1)) → 21(R(x1))
21(c(0(x1))) → 01(0(x1))
21(c(1(x0))) → 01(L(3(x0)))
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 01(L(x1)) → 21(R(x1)) at position [0] we obtained the following new rules:
01(L(b(x0))) → 21(c(1(b(x0))))
01(L(2(x0))) → 21(2(R(x0)))
01(L(1(x0))) → 21(L(3(x0)))
01(L(3(x0))) → 21(3(R(x0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
21(L(x1)) → 21(x1)
01(L(3(x0))) → 21(3(R(x0)))
01(L(b(x0))) → 21(c(1(b(x0))))
21(c(0(x1))) → 01(0(x1))
01(L(2(x0))) → 21(2(R(x0)))
21(c(1(x0))) → 01(L(3(x0)))
01(L(1(x0))) → 21(L(3(x0)))
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(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
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
21(L(x1)) → 21(x1)
01(L(3(x0))) → 21(3(R(x0)))
21(c(0(x1))) → 01(0(x1))
01(L(2(x0))) → 21(2(R(x0)))
21(c(1(x0))) → 01(L(3(x0)))
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 21(c(0(x1))) → 01(0(x1)) at position [0] we obtained the following new rules:
21(c(0(L(x0)))) → 01(2(R(x0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
21(L(x1)) → 21(x1)
21(c(0(L(x0)))) → 01(2(R(x0)))
01(L(3(x0))) → 21(3(R(x0)))
01(L(2(x0))) → 21(2(R(x0)))
21(c(1(x0))) → 01(L(3(x0)))
The TRS R consists of the following rules:
R(2(x1)) → 2(R(x1))
R(3(x1)) → 3(R(x1))
R(1(x1)) → L(3(x1))
3(L(x1)) → L(3(x1))
2(L(x1)) → L(2(x1))
0(L(x1)) → 2(R(x1))
R(b(x1)) → c(1(b(x1)))
3(c(x1)) → c(1(x1))
2(c(1(x1))) → c(0(R(1(x1))))
2(c(0(x1))) → c(0(0(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.