Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
d(c(x1)) → d(a(x1))
a(c(x1)) → c(a(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(c(x1))
a(b(x1)) → b(a(x1))
d(c(x1)) → d(a(x1))
a(c(x1)) → c(a(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:
D(c(x1)) → A(x1)
A(c(x1)) → A(x1)
D(c(x1)) → D(a(x1))
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
d(c(x1)) → d(a(x1))
a(c(x1)) → c(a(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:
D(c(x1)) → A(x1)
A(c(x1)) → A(x1)
D(c(x1)) → D(a(x1))
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
d(c(x1)) → d(a(x1))
a(c(x1)) → c(a(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(x1)) → A(x1)
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
d(c(x1)) → d(a(x1))
a(c(x1)) → c(a(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(x1)) → A(x1)
A(b(x1)) → A(x1)
R is empty.
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(x1)) → A(x1)
A(b(x1)) → A(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:
A(c(x1)) → A(x1)
A(b(x1)) → A(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(b(x1)) = 2·x1
POL(c(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(c(x1)) → D(a(x1))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
d(c(x1)) → d(a(x1))
a(c(x1)) → c(a(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(c(x1)) → D(a(x1))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
a(c(x1)) → c(a(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
D(c(x1)) → D(a(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:
POL( c(x1) ) = x1 + 1
POL( b(x1) ) = 0
POL( D(x1) ) = x1 + 1
POL( a(x1) ) = x1
The following usable rules [17] were oriented:
a(c(x1)) → c(a(x1))
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
a(c(x1)) → c(a(x1))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(c(x1)) → D(a(x1))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
a(c(x1)) → c(a(x1))
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(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
d(c(x1)) → d(a(x1))
a(c(x1)) → c(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(a(x)) → a(b(x))
c(d(x)) → a(d(x))
c(a(x)) → a(c(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) → c(b(x))
b(a(x)) → a(b(x))
c(d(x)) → a(d(x))
c(a(x)) → a(c(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
a(b(x1)) → b(a(x1))
d(c(x1)) → d(a(x1))
a(c(x1)) → c(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(a(x)) → a(b(x))
c(d(x)) → a(d(x))
c(a(x)) → a(c(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) → c(b(x))
b(a(x)) → a(b(x))
c(d(x)) → a(d(x))
c(a(x)) → a(c(x))
Q is empty.