YES
Termination w.r.t. Q proof of ../tpdb/SRS/Hofbauer/touzet.srs
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(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(x1) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(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(a(x1))
A(b(x1)) → B(b(b(x1)))
B(a(x1)) → A(a(a(x1)))
A(b(x1)) → B(b(x1))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
B(a(x1)) → A(a(x1))
A(b(x1)) → B(b(b(x1)))
B(a(x1)) → A(a(a(x1)))
A(b(x1)) → B(b(x1))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → B(b(b(x1))) at position [0] we obtained the following new rules:
A(b(a(x0))) → B(b(a(a(a(x0)))))
A(b(y0)) → B(b(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
B(a(x1)) → A(a(x1))
B(a(x1)) → A(a(a(x1)))
A(b(a(x0))) → B(b(a(a(a(x0)))))
A(b(x1)) → B(b(x1))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → B(b(x1)) at position [0] we obtained the following new rules:
A(b(x0)) → B(x0)
A(b(a(x0))) → B(a(a(a(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
B(a(x1)) → A(a(x1))
A(b(a(x0))) → B(a(a(a(x0))))
A(b(x0)) → B(x0)
B(a(x1)) → A(a(a(x1)))
A(b(a(x0))) → B(b(a(a(a(x0)))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(x1)) → A(a(a(x1))) at position [0] we obtained the following new rules:
B(a(b(x0))) → A(a(b(b(b(x0)))))
B(a(y0)) → A(a(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
B(a(x1)) → A(a(x1))
A(b(a(x0))) → B(a(a(a(x0))))
B(a(b(x0))) → A(a(b(b(b(x0)))))
A(b(x0)) → B(x0)
A(b(a(x0))) → B(b(a(a(a(x0)))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(x1)) → A(a(x1)) at position [0] we obtained the following new rules:
B(a(x0)) → A(x0)
B(a(b(x0))) → A(b(b(b(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
Q DP problem:
The TRS P consists of the following rules:
A(b(a(x0))) → B(a(a(a(x0))))
A(b(x0)) → B(x0)
B(a(b(x0))) → A(a(b(b(b(x0)))))
B(a(b(x0))) → A(b(b(b(x0))))
B(a(x0)) → A(x0)
A(b(a(x0))) → B(b(a(a(a(x0)))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(x1) → x1
A(b(a(x0))) → B(a(a(a(x0))))
A(b(x0)) → B(x0)
B(a(b(x0))) → A(a(b(b(b(x0)))))
B(a(b(x0))) → A(b(b(b(x0))))
B(a(x0)) → A(x0)
A(b(a(x0))) → B(b(a(a(a(x0)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → b(b(b(x1)))
b(a(x1)) → a(a(a(x1)))
a(x1) → x1
b(x1) → x1
A(b(a(x0))) → B(a(a(a(x0))))
A(b(x0)) → B(x0)
B(a(b(x0))) → A(a(b(b(b(x0)))))
B(a(b(x0))) → A(b(b(b(x0))))
B(a(x0)) → A(x0)
A(b(a(x0))) → B(b(a(a(a(x0)))))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → b(b(b(x)))
a(b(x)) → a(a(a(x)))
a(x) → x
b(x) → x
a(b(A(x))) → a(a(a(B(x))))
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(B(x))) → b(b(b(A(x))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(b(B(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(x)) → b(b(b(x)))
a(b(x)) → a(a(a(x)))
a(x) → x
b(x) → x
a(b(A(x))) → a(a(a(B(x))))
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(B(x))) → b(b(b(A(x))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(b(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(A(x))) → A1(B(x))
A1(b(A(x))) → A1(a(B(x)))
B1(a(B(x))) → A1(A(x))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(b(x))
A1(b(x)) → A1(a(x))
A1(b(A(x))) → B1(B(x))
A1(b(A(x))) → A1(a(b(B(x))))
B1(a(B(x))) → B1(a(A(x)))
B1(a(B(x))) → B1(b(b(a(A(x)))))
A1(b(A(x))) → A1(a(a(b(B(x)))))
B1(a(B(x))) → B1(b(a(A(x))))
A1(b(x)) → A1(a(a(x)))
A1(b(A(x))) → A1(a(a(B(x))))
B1(a(B(x))) → B1(A(x))
A1(b(A(x))) → A1(b(B(x)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(b(b(x)))
B1(a(B(x))) → B1(b(b(A(x))))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
b(a(x)) → b(b(b(x)))
a(b(x)) → a(a(a(x)))
a(x) → x
b(x) → x
a(b(A(x))) → a(a(a(B(x))))
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(B(x))) → b(b(b(A(x))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(b(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(B(x))
A1(b(A(x))) → A1(a(B(x)))
B1(a(B(x))) → A1(A(x))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(b(x))
A1(b(x)) → A1(a(x))
A1(b(A(x))) → B1(B(x))
A1(b(A(x))) → A1(a(b(B(x))))
B1(a(B(x))) → B1(a(A(x)))
B1(a(B(x))) → B1(b(b(a(A(x)))))
A1(b(A(x))) → A1(a(a(b(B(x)))))
B1(a(B(x))) → B1(b(a(A(x))))
A1(b(x)) → A1(a(a(x)))
A1(b(A(x))) → A1(a(a(B(x))))
B1(a(B(x))) → B1(A(x))
A1(b(A(x))) → A1(b(B(x)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(b(b(x)))
B1(a(B(x))) → B1(b(b(A(x))))
A1(b(x)) → A1(x)
The TRS R consists of the following rules:
b(a(x)) → b(b(b(x)))
a(b(x)) → a(a(a(x)))
a(x) → x
b(x) → x
a(b(A(x))) → a(a(a(B(x))))
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(B(x))) → b(b(b(A(x))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(b(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 2 SCCs with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(x)))
A1(b(A(x))) → A1(a(B(x)))
A1(b(x)) → A1(a(x))
A1(b(A(x))) → A1(b(B(x)))
A1(b(x)) → A1(x)
A1(b(A(x))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(x)))
a(b(x)) → a(a(a(x)))
a(x) → x
b(x) → x
a(b(A(x))) → a(a(a(B(x))))
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(B(x))) → b(b(b(A(x))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(b(B(x)))))
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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(x)) → A1(a(a(x)))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(A(x))) → A1(a(B(x)))
A1(b(x)) → A1(a(x))
A1(b(A(x))) → A1(b(B(x)))
A1(b(x)) → A1(x)
A1(b(A(x))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
a(b(A(x))) → a(a(a(B(x))))
a(b(A(x))) → a(a(a(b(B(x)))))
a(b(x)) → a(a(a(x)))
a(x) → x
a(B(x)) → A(x)
b(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(A(x))) → A1(a(B(x))) at position [0] we obtained the following new rules:
A1(b(A(y0))) → A1(B(y0))
A1(b(A(x0))) → A1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(y0))) → A1(B(y0))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(x)))
A1(b(A(x0))) → A1(A(x0))
A1(b(x)) → A1(a(x))
A1(b(A(x))) → A1(b(B(x)))
A1(b(x)) → A1(x)
A1(b(A(x))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
a(b(A(x))) → a(a(a(B(x))))
a(b(A(x))) → a(a(a(b(B(x)))))
a(b(x)) → a(a(a(x)))
a(x) → x
a(B(x)) → A(x)
b(x) → 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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(x)))
A1(b(x)) → A1(a(x))
A1(b(A(x))) → A1(b(B(x)))
A1(b(x)) → A1(x)
A1(b(A(x))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
a(b(A(x))) → a(a(a(B(x))))
a(b(A(x))) → a(a(a(b(B(x)))))
a(b(x)) → a(a(a(x)))
a(x) → x
a(B(x)) → A(x)
b(x) → 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:
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(x)))
A1(b(x)) → A1(a(x))
A1(b(x)) → A1(x)
Strictly oriented rules of the TRS R:
a(b(A(x))) → a(a(a(B(x))))
a(b(x)) → a(a(a(x)))
b(x) → x
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2 + 2·x1
POL(A1(x1)) = 2·x1
POL(B(x1)) = 2 + 2·x1
POL(a(x1)) = x1
POL(b(x1)) = 2 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(b(B(x)))
A1(b(A(x))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
a(b(A(x))) → a(a(a(b(B(x)))))
a(x) → x
a(B(x)) → A(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
a(b(A(x))) → a(a(a(b(B(x)))))
a(x) → x
a(B(x)) → A(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:
A1(b(A(x))) → A1(a(b(B(x))))
Strictly oriented rules of the TRS R:
a(x) → x
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2 + x1
POL(A1(x1)) = x1
POL(B(x1)) = x1
POL(a(x1)) = 2 + x1
POL(b(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(a(a(b(B(x)))))
The TRS R consists of the following rules:
a(b(A(x))) → a(a(a(b(B(x)))))
a(B(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(x)) → A1(a(a(x)))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(A(x))) → A1(a(B(x)))
A1(b(x)) → A1(a(x))
A1(b(A(x))) → A1(b(B(x)))
A1(b(x)) → A1(x)
A1(b(A(x))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
a(b(A(x))) → a(a(a(B(x))))
a(b(A(x))) → a(a(a(b(B(x)))))
a(b(x)) → a(a(a(x)))
a(x) → x
a(B(x)) → A(x)
b(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(B(x))) → B1(b(a(A(x))))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(b(x))
B1(a(x)) → B1(x)
B1(a(x)) → B1(b(b(x)))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → B1(a(A(x)))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(x)))
a(b(x)) → a(a(a(x)))
a(x) → x
b(x) → x
a(b(A(x))) → a(a(a(B(x))))
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(B(x))) → b(b(b(A(x))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(b(B(x)))))
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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(B(x))) → B1(b(a(A(x))))
B1(a(x)) → B1(b(x))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(b(b(x)))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → B1(a(A(x)))
The TRS R consists of the following rules:
b(x) → x
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(A(x))))
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(x)) → b(b(b(x)))
a(x) → 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:
B1(a(x)) → B1(b(x))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(b(b(x)))
B1(a(B(x))) → B1(b(b(A(x))))
Strictly oriented rules of the TRS R:
b(a(B(x))) → b(b(b(A(x))))
b(a(x)) → b(b(b(x)))
a(x) → x
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(B(x1)) = x1
POL(B1(x1)) = x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(B(x))) → B1(b(a(A(x))))
B1(a(B(x))) → B1(a(A(x)))
The TRS R consists of the following rules:
b(x) → x
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(B(x))) → B1(b(a(A(x))))
The TRS R consists of the following rules:
b(x) → x
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(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:
B1(a(B(x))) → B1(b(a(A(x))))
Strictly oriented rules of the TRS R:
b(x) → x
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(B(x1)) = 2 + 2·x1
POL(B1(x1)) = 2·x1
POL(a(x1)) = 2·x1
POL(b(x1)) = 2 + x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(B(x))) → B1(b(b(a(A(x)))))
The TRS R consists of the following rules:
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ 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:
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(B(x))) → B1(b(a(A(x))))
B1(a(x)) → B1(b(x))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(x)
B1(a(x)) → B1(b(b(x)))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → B1(a(A(x)))
The TRS R consists of the following rules:
b(x) → x
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(A(x))))
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(x)) → b(b(b(x)))
a(x) → 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)) → b(b(b(x)))
a(b(x)) → a(a(a(x)))
a(x) → x
b(x) → x
a(b(A(x))) → a(a(a(B(x))))
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(B(x))) → b(b(b(A(x))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(b(B(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → b(b(b(x)))
b(a(x)) → a(a(a(x)))
a(x) → x
b(x) → x
A(b(a(x))) → B(a(a(a(x))))
A(b(x)) → B(x)
B(a(b(x))) → A(a(b(b(b(x)))))
B(a(b(x))) → A(b(b(b(x))))
B(a(x)) → A(x)
A(b(a(x))) → B(b(a(a(a(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(b(x)))
b(a(x)) → a(a(a(x)))
a(x) → x
b(x) → x
A(b(a(x))) → B(a(a(a(x))))
A(b(x)) → B(x)
B(a(b(x))) → A(a(b(b(b(x)))))
B(a(b(x))) → A(b(b(b(x))))
B(a(x)) → A(x)
A(b(a(x))) → B(b(a(a(a(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(x)) → b(b(b(x)))
a(b(x)) → a(a(a(x)))
a(x) → x
b(x) → x
a(b(A(x))) → a(a(a(B(x))))
b(A(x)) → B(x)
b(a(B(x))) → b(b(b(a(A(x)))))
b(a(B(x))) → b(b(b(A(x))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(b(B(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → b(b(b(x)))
b(a(x)) → a(a(a(x)))
a(x) → x
b(x) → x
A(b(a(x))) → B(a(a(a(x))))
A(b(x)) → B(x)
B(a(b(x))) → A(a(b(b(b(x)))))
B(a(b(x))) → A(b(b(b(x))))
B(a(x)) → A(x)
A(b(a(x))) → B(b(a(a(a(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(b(x)))
b(a(x)) → a(a(a(x)))
a(x) → x
b(x) → x
A(b(a(x))) → B(a(a(a(x))))
A(b(x)) → B(x)
B(a(b(x))) → A(a(b(b(b(x)))))
B(a(b(x))) → A(b(b(b(x))))
B(a(x)) → A(x)
A(b(a(x))) → B(b(a(a(a(x)))))
Q is empty.