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:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(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:
A(a(a(x1))) → B(a(a(x1)))
B(c(x1)) → A(x1)
A(c(x1)) → B(x1)
A(a(x1)) → A(b(a(x1)))
B(b(b(x1))) → A(x1)
A(c(x1)) → B(b(x1))
B(c(x1)) → A(a(x1))
A(a(x1)) → B(a(x1))
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(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:
A(a(a(x1))) → B(a(a(x1)))
B(c(x1)) → A(x1)
A(c(x1)) → B(x1)
A(a(x1)) → A(b(a(x1)))
B(b(b(x1))) → A(x1)
A(c(x1)) → B(b(x1))
B(c(x1)) → A(a(x1))
A(a(x1)) → B(a(x1))
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(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:
B(c(x1)) → A(x1)
A(c(x1)) → B(x1)
A(c(x1)) → B(b(x1))
B(c(x1)) → A(a(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(B(x1)) = 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:
A(a(a(x1))) → B(a(a(x1)))
A(a(x1)) → A(b(a(x1)))
B(b(b(x1))) → A(x1)
A(a(x1)) → B(a(x1))
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(a(x1))) → B(a(a(x1))) at position [0] we obtained the following new rules:
A(a(a(a(x0)))) → B(b(a(a(x0))))
A(a(a(c(x0)))) → B(a(c(b(b(x0)))))
A(a(a(x0))) → B(a(b(a(x0))))
A(a(a(a(a(x0))))) → B(a(b(a(a(x0)))))
A(a(a(a(x0)))) → B(a(a(b(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(x0))) → B(a(b(a(x0))))
A(a(a(a(a(x0))))) → B(a(b(a(a(x0)))))
A(a(x1)) → A(b(a(x1)))
A(a(a(a(x0)))) → B(b(a(a(x0))))
A(a(a(c(x0)))) → B(a(c(b(b(x0)))))
B(b(b(x1))) → A(x1)
A(a(x1)) → B(a(x1))
A(a(a(a(x0)))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(x1)) → A(b(a(x1))) at position [0] we obtained the following new rules:
A(a(a(a(x0)))) → A(b(b(a(a(x0)))))
A(a(c(x0))) → A(b(c(b(b(x0)))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(a(x0)))) → A(b(b(a(a(x0)))))
A(a(c(x0))) → A(b(c(b(b(x0)))))
A(a(a(x0))) → B(a(b(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(a(a(a(x0))))) → B(a(b(a(a(x0)))))
A(a(a(a(x0)))) → B(b(a(a(x0))))
B(b(b(x1))) → A(x1)
A(a(a(c(x0)))) → B(a(c(b(b(x0)))))
A(a(x1)) → B(a(x1))
A(a(a(a(x0)))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(x1)) → B(a(x1)) at position [0] we obtained the following new rules:
A(a(a(a(x0)))) → B(b(a(a(x0))))
A(a(c(x0))) → B(c(b(b(x0))))
A(a(a(x0))) → B(a(b(a(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(a(x0)))) → A(b(b(a(a(x0)))))
A(a(a(x0))) → B(a(b(a(x0))))
A(a(c(x0))) → A(b(c(b(b(x0)))))
A(a(a(a(a(x0))))) → B(a(b(a(a(x0)))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(a(a(x0)))) → B(b(a(a(x0))))
A(a(c(x0))) → B(c(b(b(x0))))
A(a(a(c(x0)))) → B(a(c(b(b(x0)))))
B(b(b(x1))) → A(x1)
A(a(a(a(x0)))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(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 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(a(x0)))) → A(b(b(a(a(x0)))))
A(a(c(x0))) → A(b(c(b(b(x0)))))
A(a(a(x0))) → B(a(b(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(a(a(a(x0))))) → B(a(b(a(a(x0)))))
A(a(a(a(x0)))) → B(b(a(a(x0))))
A(a(a(c(x0)))) → B(a(c(b(b(x0)))))
B(b(b(x1))) → A(x1)
A(a(a(a(x0)))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(x1)))
A(a(a(a(x0)))) → A(b(b(a(a(x0)))))
A(a(c(x0))) → A(b(c(b(b(x0)))))
A(a(a(x0))) → B(a(b(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(a(a(a(x0))))) → B(a(b(a(a(x0)))))
A(a(a(a(x0)))) → B(b(a(a(x0))))
A(a(a(c(x0)))) → B(a(c(b(b(x0)))))
B(b(b(x1))) → A(x1)
A(a(a(a(x0)))) → B(a(a(b(a(x0)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(x1)))
A(a(a(a(x0)))) → A(b(b(a(a(x0)))))
A(a(c(x0))) → A(b(c(b(b(x0)))))
A(a(a(x0))) → B(a(b(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(a(a(a(x0))))) → B(a(b(a(a(x0)))))
A(a(a(a(x0)))) → B(b(a(a(x0))))
A(a(a(c(x0)))) → B(a(c(b(b(x0)))))
B(b(b(x1))) → A(x1)
A(a(a(a(x0)))) → B(a(a(b(a(x0)))))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
c(a(A(x))) → b(b(c(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(a(b(B(x))))
c(a(a(A(x)))) → b(b(c(a(B(x)))))
b(b(B(x))) → A(x)
a(a(a(A(x)))) → a(b(a(a(B(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
c(a(A(x))) → b(b(c(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(a(b(B(x))))
c(a(a(A(x)))) → b(b(c(a(B(x)))))
b(b(B(x))) → A(x)
a(a(a(A(x)))) → a(b(a(a(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(a(a(a(A(x))))) → A1(b(a(B(x))))
A1(a(A(x))) → B1(a(B(x)))
A1(a(a(A(x)))) → B1(B(x))
A1(a(a(A(x)))) → A1(B(x))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
B1(b(b(x))) → A1(x)
C(a(A(x))) → B1(A(x))
C(a(A(x))) → B1(b(c(b(A(x)))))
A1(a(A(x))) → A1(b(a(b(A(x)))))
A1(a(a(A(x)))) → A1(a(B(x)))
C(a(a(A(x)))) → B1(b(c(a(B(x)))))
A1(a(a(A(x)))) → A1(b(B(x)))
A1(a(x)) → A1(b(a(x)))
A1(a(a(A(x)))) → A1(a(b(B(x))))
A1(a(a(a(A(x))))) → A1(a(b(a(B(x)))))
A1(a(A(x))) → A1(b(A(x)))
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(A(x)))) → A1(a(b(b(A(x)))))
C(a(A(x))) → B1(c(b(A(x))))
C(a(A(x))) → C(b(A(x)))
A1(a(a(A(x)))) → A1(b(b(A(x))))
A1(a(a(a(A(x))))) → B1(a(B(x)))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
A1(a(a(A(x)))) → B1(b(A(x)))
A1(a(A(x))) → A1(b(a(B(x))))
A1(a(a(A(x)))) → B1(a(a(B(x))))
C(a(x)) → C(x)
C(a(a(A(x)))) → A1(B(x))
C(a(x)) → B1(b(c(x)))
A1(a(A(x))) → B1(a(b(A(x))))
C(a(a(A(x)))) → B1(c(a(B(x))))
A1(a(A(x))) → B1(A(x))
C(b(x)) → C(x)
A1(a(a(A(x)))) → B1(A(x))
C(b(x)) → A1(a(c(x)))
C(a(a(A(x)))) → C(a(B(x)))
C(a(x)) → B1(c(x))
A1(a(a(a(A(x))))) → A1(B(x))
C(b(x)) → A1(c(x))
A1(a(x)) → B1(a(x))
A1(a(A(x))) → A1(B(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
c(a(A(x))) → b(b(c(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(a(b(B(x))))
c(a(a(A(x)))) → b(b(c(a(B(x)))))
b(b(B(x))) → A(x)
a(a(a(A(x)))) → a(b(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(a(a(a(A(x))))) → A1(b(a(B(x))))
A1(a(A(x))) → B1(a(B(x)))
A1(a(a(A(x)))) → B1(B(x))
A1(a(a(A(x)))) → A1(B(x))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
B1(b(b(x))) → A1(x)
C(a(A(x))) → B1(A(x))
C(a(A(x))) → B1(b(c(b(A(x)))))
A1(a(A(x))) → A1(b(a(b(A(x)))))
A1(a(a(A(x)))) → A1(a(B(x)))
C(a(a(A(x)))) → B1(b(c(a(B(x)))))
A1(a(a(A(x)))) → A1(b(B(x)))
A1(a(x)) → A1(b(a(x)))
A1(a(a(A(x)))) → A1(a(b(B(x))))
A1(a(a(a(A(x))))) → A1(a(b(a(B(x)))))
A1(a(A(x))) → A1(b(A(x)))
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(A(x)))) → A1(a(b(b(A(x)))))
C(a(A(x))) → B1(c(b(A(x))))
C(a(A(x))) → C(b(A(x)))
A1(a(a(A(x)))) → A1(b(b(A(x))))
A1(a(a(a(A(x))))) → B1(a(B(x)))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
A1(a(a(A(x)))) → B1(b(A(x)))
A1(a(A(x))) → A1(b(a(B(x))))
A1(a(a(A(x)))) → B1(a(a(B(x))))
C(a(x)) → C(x)
C(a(a(A(x)))) → A1(B(x))
C(a(x)) → B1(b(c(x)))
A1(a(A(x))) → B1(a(b(A(x))))
C(a(a(A(x)))) → B1(c(a(B(x))))
A1(a(A(x))) → B1(A(x))
C(b(x)) → C(x)
A1(a(a(A(x)))) → B1(A(x))
C(b(x)) → A1(a(c(x)))
C(a(a(A(x)))) → C(a(B(x)))
C(a(x)) → B1(c(x))
A1(a(a(a(A(x))))) → A1(B(x))
C(b(x)) → A1(c(x))
A1(a(x)) → B1(a(x))
A1(a(A(x))) → A1(B(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
c(a(A(x))) → b(b(c(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(a(b(B(x))))
c(a(a(A(x)))) → b(b(c(a(B(x)))))
b(b(B(x))) → A(x)
a(a(a(A(x)))) → a(b(a(a(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 26 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(A(x)))) → B1(a(a(B(x))))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
A1(a(a(A(x)))) → A1(a(b(b(A(x)))))
B1(b(b(x))) → A1(x)
A1(a(a(A(x)))) → A1(a(B(x)))
A1(a(x)) → B1(a(x))
A1(a(x)) → A1(b(a(x)))
A1(a(a(A(x)))) → A1(a(b(B(x))))
A1(a(a(a(A(x))))) → A1(a(b(a(B(x)))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
c(a(A(x))) → b(b(c(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(a(b(B(x))))
c(a(a(A(x)))) → b(b(c(a(B(x)))))
b(b(B(x))) → A(x)
a(a(a(A(x)))) → a(b(a(a(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(A(x)))) → B1(a(a(B(x))))
B1(b(b(x))) → A1(x)
A1(a(a(A(x)))) → A1(a(b(b(A(x)))))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
A1(a(x)) → B1(a(x))
A1(a(a(A(x)))) → A1(a(B(x)))
A1(a(x)) → A1(b(a(x)))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
A1(a(a(a(A(x))))) → A1(a(b(a(B(x)))))
A1(a(a(A(x)))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
A1(a(a(A(x)))) → A1(a(b(b(A(x)))))
B1(b(b(x))) → A1(x)
A1(a(a(A(x)))) → A1(a(B(x)))
A1(a(x)) → A1(b(a(x)))
A1(a(a(A(x)))) → A1(a(b(B(x))))
A1(a(a(a(A(x))))) → A1(a(b(a(B(x)))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(A(x)))) → B1(a(a(B(x))))
B1(b(b(x))) → A1(x)
A1(a(a(A(x)))) → A1(a(b(b(A(x)))))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
A1(a(x)) → B1(a(x))
A1(a(a(A(x)))) → A1(a(B(x)))
A1(a(x)) → A1(b(a(x)))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
A1(a(a(a(A(x))))) → A1(a(b(a(B(x)))))
A1(a(a(A(x)))) → A1(a(b(B(x))))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
A1(a(a(A(x)))) → A1(a(b(b(A(x)))))
B1(b(b(x))) → A1(x)
A1(a(a(A(x)))) → A1(a(B(x)))
A1(a(x)) → A1(b(a(x)))
A1(a(a(A(x)))) → A1(a(b(B(x))))
A1(a(a(a(A(x))))) → A1(a(b(a(B(x)))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(x)) → A1(b(a(x))) at position [0] we obtained the following new rules:
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
A1(a(a(A(x0)))) → A1(b(a(b(a(b(A(x0)))))))
A1(a(a(a(A(x0))))) → A1(b(a(a(b(b(A(x0)))))))
A1(a(a(a(A(x0))))) → A1(b(a(a(b(B(x0))))))
A1(a(a(A(x0)))) → A1(b(a(b(a(B(x0))))))
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ 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:
A1(a(a(A(x0)))) → A1(b(a(b(a(b(A(x0)))))))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
B1(b(b(x))) → A1(x)
A1(a(a(A(x)))) → A1(a(B(x)))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(A(x)))) → A1(a(b(B(x))))
A1(a(a(a(A(x))))) → A1(a(b(a(B(x)))))
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
A1(a(a(A(x)))) → A1(a(b(b(A(x)))))
A1(a(a(a(A(x0))))) → A1(b(a(a(b(b(A(x0)))))))
A1(a(a(a(A(x0))))) → A1(b(a(a(b(B(x0))))))
A1(a(a(A(x0)))) → A1(b(a(b(a(B(x0))))))
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
A1(a(a(a(A(x0))))) → A1(b(a(a(b(b(A(x0)))))))
A1(a(a(A(x)))) → A1(b(a(a(B(x)))))
B1(b(b(x))) → A1(x)
A1(a(a(a(A(x0))))) → A1(b(a(a(b(B(x0))))))
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(A(x)))) → A1(b(a(a(B(x))))) at position [0] we obtained the following new rules:
A1(a(a(A(y0)))) → A1(b(a(b(a(B(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ 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:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(a(A(x0))))) → A1(b(a(a(b(b(A(x0)))))))
A1(a(a(a(A(x0))))) → A1(b(a(a(b(B(x0))))))
A1(a(a(A(y0)))) → A1(b(a(b(a(B(y0))))))
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
A1(a(a(a(A(x0))))) → A1(b(a(a(b(b(A(x0)))))))
B1(b(b(x))) → A1(x)
A1(a(a(a(A(x0))))) → A1(b(a(a(b(B(x0))))))
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(a(A(x0))))) → A1(b(a(a(b(b(A(x0))))))) at position [0] we obtained the following new rules:
A1(a(a(a(A(y0))))) → A1(b(a(b(a(b(b(A(y0))))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(a(a(a(A(y0))))) → A1(b(a(b(a(b(b(A(y0))))))))
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(a(A(x0))))) → A1(b(a(a(b(B(x0))))))
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(a(A(x0))))) → A1(b(a(a(b(B(x0))))))
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(a(A(x0))))) → A1(b(a(a(b(B(x0)))))) at position [0] we obtained the following new rules:
A1(a(a(a(A(y0))))) → A1(b(a(b(a(b(B(y0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(a(a(a(A(y0))))) → A1(b(a(b(a(b(B(y0)))))))
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(a(A(x0))))) → A1(b(a(b(a(a(B(x0))))))) at position [0] we obtained the following new rules:
A1(a(a(a(A(y0))))) → A1(b(a(b(a(b(a(B(y0))))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
A1(a(a(a(A(y0))))) → A1(b(a(b(a(b(a(B(y0))))))))
B1(b(b(x))) → A1(x)
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0)))))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(a(a(A(x0)))))) → A1(b(a(a(b(a(B(x0))))))) at position [0] we obtained the following new rules:
A1(a(a(a(a(A(y0)))))) → A1(b(a(b(a(b(a(B(y0))))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(a(a(A(y0)))))) → A1(b(a(b(a(b(a(B(y0))))))))
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → A1(a(b(x)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(x))) → A1(a(b(x))) at position [0] we obtained the following new rules:
A1(a(a(b(B(x0))))) → A1(a(A(x0)))
A1(a(a(b(b(x0))))) → A1(a(a(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(a(a(b(b(x0))))) → A1(a(a(x0)))
A1(a(a(b(B(x0))))) → A1(a(A(x0)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(b(x0))))) → A1(a(a(x0)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(x))) → A1(b(x)) at position [0] we obtained the following new rules:
A1(a(a(b(b(x0))))) → A1(a(x0))
A1(a(a(b(B(x0))))) → A1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(a(a(b(b(x0))))) → A1(a(x0))
A1(a(a(b(b(x0))))) → A1(a(a(x0)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
A1(a(a(b(B(x0))))) → A1(A(x0))
B1(b(b(x))) → A1(x)
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(b(x0))))) → A1(a(x0))
A1(a(a(b(b(x0))))) → A1(a(a(x0)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A1(a(a(x))) → B1(x) we obtained the following new rules:
A1(a(a(b(b(y_0))))) → B1(b(b(y_0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(b(x0))))) → A1(a(x0))
A1(a(a(b(b(x0))))) → A1(a(a(x0)))
A1(a(a(a(x0)))) → A1(b(a(a(b(x0)))))
B1(b(b(x))) → A1(x)
A1(a(a(x0))) → A1(b(a(b(a(x0)))))
A1(a(a(b(b(y_0))))) → B1(b(b(y_0)))
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
b(b(B(x))) → A(x)
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(b(a(a(B(x)))))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(B(x))))
a(a(x)) → a(b(a(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.A1: 0
B: 0
a: 0
A: 0
B1: 0
b: 1
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
A1.0(a.0(a.1(b.1(b.1(y_0))))) → B1.1(b.1(b.1(y_0)))
A1.0(a.0(a.1(b.1(b.0(y_0))))) → B1.0(b.1(b.0(y_0)))
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.1(x0))
A1.0(a.0(a.0(x0))) → A1.0(b.0(a.1(b.0(a.0(x0)))))
A1.0(a.0(a.0(a.0(x0)))) → A1.0(b.0(a.0(a.1(b.0(x0)))))
A1.0(a.0(a.1(x0))) → A1.1(b.0(a.1(b.0(a.1(x0)))))
A1.0(a.0(a.1(b.1(b.1(y_0))))) → B1.0(b.1(b.1(y_0)))
A1.0(a.0(a.0(a.0(x0)))) → A1.1(b.0(a.0(a.1(b.0(x0)))))
B1.1(b.1(b.1(x))) → A1.1(x)
A1.0(a.0(a.1(b.1(b.0(y_0))))) → B1.1(b.1(b.0(y_0)))
B1.1(b.1(b.1(x))) → A1.0(x)
A1.0(a.0(a.0(x0))) → A1.1(b.0(a.1(b.0(a.0(x0)))))
A1.0(a.0(a.0(a.1(x0)))) → A1.0(b.0(a.0(a.1(b.1(x0)))))
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.0(a.1(x0)))
B1.1(b.1(b.0(x))) → A1.0(x)
A1.0(a.0(a.1(x0))) → A1.0(b.0(a.1(b.0(a.1(x0)))))
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(x0))
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(a.0(x0)))
A1.0(a.0(a.0(a.1(x0)))) → A1.1(b.0(a.0(a.1(b.1(x0)))))
The TRS R consists of the following rules:
b.1(b.1(b.0(x))) → a.0(x)
a.0(a.0(A.0(x))) → a.1(b.0(a.0(B.0(x))))
A.1(x0) → A.0(x0)
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.0(B.0(x))))
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.0(B.1(x))))
b.1(x0) → b.0(x0)
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.1(b.0(A.1(x)))))
a.0(a.1(x)) → a.1(b.0(a.1(x)))
a.0(a.0(a.0(a.0(A.1(x))))) → a.0(a.1(b.0(a.0(B.1(x)))))
a.0(a.0(a.0(a.0(A.0(x))))) → a.0(a.1(b.0(a.0(B.0(x)))))
a.1(x0) → a.0(x0)
b.1(b.1(b.1(x))) → a.1(x)
a.0(a.0(x)) → a.1(b.0(a.0(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.1(b.0(A.1(x)))))
a.0(a.0(A.0(x))) → a.1(b.0(a.1(b.0(A.0(x)))))
a.0(a.0(a.0(A.0(x)))) → a.1(b.0(a.0(a.0(B.0(x)))))
B.1(x0) → B.0(x0)
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.1(b.0(A.0(x)))))
a.0(a.0(a.0(x))) → a.0(a.1(b.0(x)))
b.1(b.0(B.0(x))) → A.0(x)
a.0(a.0(a.1(x))) → a.0(a.1(b.1(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.0(B.1(x))))
a.0(a.0(a.0(A.1(x)))) → a.1(b.0(a.0(a.0(B.1(x)))))
b.1(b.0(B.1(x))) → A.1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1.0(a.0(a.1(b.1(b.1(y_0))))) → B1.1(b.1(b.1(y_0)))
A1.0(a.0(a.1(b.1(b.0(y_0))))) → B1.0(b.1(b.0(y_0)))
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.1(x0))
A1.0(a.0(a.0(x0))) → A1.0(b.0(a.1(b.0(a.0(x0)))))
A1.0(a.0(a.0(a.0(x0)))) → A1.0(b.0(a.0(a.1(b.0(x0)))))
A1.0(a.0(a.1(x0))) → A1.1(b.0(a.1(b.0(a.1(x0)))))
A1.0(a.0(a.1(b.1(b.1(y_0))))) → B1.0(b.1(b.1(y_0)))
A1.0(a.0(a.0(a.0(x0)))) → A1.1(b.0(a.0(a.1(b.0(x0)))))
B1.1(b.1(b.1(x))) → A1.1(x)
A1.0(a.0(a.1(b.1(b.0(y_0))))) → B1.1(b.1(b.0(y_0)))
B1.1(b.1(b.1(x))) → A1.0(x)
A1.0(a.0(a.0(x0))) → A1.1(b.0(a.1(b.0(a.0(x0)))))
A1.0(a.0(a.0(a.1(x0)))) → A1.0(b.0(a.0(a.1(b.1(x0)))))
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.0(a.1(x0)))
B1.1(b.1(b.0(x))) → A1.0(x)
A1.0(a.0(a.1(x0))) → A1.0(b.0(a.1(b.0(a.1(x0)))))
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(x0))
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(a.0(x0)))
A1.0(a.0(a.0(a.1(x0)))) → A1.1(b.0(a.0(a.1(b.1(x0)))))
The TRS R consists of the following rules:
b.1(b.1(b.0(x))) → a.0(x)
a.0(a.0(A.0(x))) → a.1(b.0(a.0(B.0(x))))
A.1(x0) → A.0(x0)
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.0(B.0(x))))
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.0(B.1(x))))
b.1(x0) → b.0(x0)
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.1(b.0(A.1(x)))))
a.0(a.1(x)) → a.1(b.0(a.1(x)))
a.0(a.0(a.0(a.0(A.1(x))))) → a.0(a.1(b.0(a.0(B.1(x)))))
a.0(a.0(a.0(a.0(A.0(x))))) → a.0(a.1(b.0(a.0(B.0(x)))))
a.1(x0) → a.0(x0)
b.1(b.1(b.1(x))) → a.1(x)
a.0(a.0(x)) → a.1(b.0(a.0(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.1(b.0(A.1(x)))))
a.0(a.0(A.0(x))) → a.1(b.0(a.1(b.0(A.0(x)))))
a.0(a.0(a.0(A.0(x)))) → a.1(b.0(a.0(a.0(B.0(x)))))
B.1(x0) → B.0(x0)
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.1(b.0(A.0(x)))))
a.0(a.0(a.0(x))) → a.0(a.1(b.0(x)))
b.1(b.0(B.0(x))) → A.0(x)
a.0(a.0(a.1(x))) → a.0(a.1(b.1(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.0(B.1(x))))
a.0(a.0(a.0(A.1(x)))) → a.1(b.0(a.0(a.0(B.1(x)))))
b.1(b.0(B.1(x))) → A.1(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 11 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1.0(a.0(a.1(b.1(b.1(y_0))))) → B1.1(b.1(b.1(y_0)))
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.1(x0))
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.0(a.1(x0)))
B1.1(b.1(b.0(x))) → A1.0(x)
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(x0))
A1.0(a.0(a.1(b.1(b.0(y_0))))) → B1.1(b.1(b.0(y_0)))
B1.1(b.1(b.1(x))) → A1.0(x)
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(a.0(x0)))
The TRS R consists of the following rules:
b.1(b.1(b.0(x))) → a.0(x)
a.0(a.0(A.0(x))) → a.1(b.0(a.0(B.0(x))))
A.1(x0) → A.0(x0)
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.0(B.0(x))))
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.0(B.1(x))))
b.1(x0) → b.0(x0)
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.1(b.0(A.1(x)))))
a.0(a.1(x)) → a.1(b.0(a.1(x)))
a.0(a.0(a.0(a.0(A.1(x))))) → a.0(a.1(b.0(a.0(B.1(x)))))
a.0(a.0(a.0(a.0(A.0(x))))) → a.0(a.1(b.0(a.0(B.0(x)))))
a.1(x0) → a.0(x0)
b.1(b.1(b.1(x))) → a.1(x)
a.0(a.0(x)) → a.1(b.0(a.0(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.1(b.0(A.1(x)))))
a.0(a.0(A.0(x))) → a.1(b.0(a.1(b.0(A.0(x)))))
a.0(a.0(a.0(A.0(x)))) → a.1(b.0(a.0(a.0(B.0(x)))))
B.1(x0) → B.0(x0)
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.1(b.0(A.0(x)))))
a.0(a.0(a.0(x))) → a.0(a.1(b.0(x)))
b.1(b.0(B.0(x))) → A.0(x)
a.0(a.0(a.1(x))) → a.0(a.1(b.1(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.0(B.1(x))))
a.0(a.0(a.0(A.1(x)))) → a.1(b.0(a.0(a.0(B.1(x)))))
b.1(b.0(B.1(x))) → A.1(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 rules of the TRS R:
A.1(x0) → A.0(x0)
B.1(x0) → B.0(x0)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(A.1(x1)) = 1 + x1
POL(A1.0(x1)) = x1
POL(B.0(x1)) = x1
POL(B.1(x1)) = 1 + x1
POL(B1.1(x1)) = x1
POL(a.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.0(a.1(x0)))
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.1(x0))
A1.0(a.0(a.1(b.1(b.1(y_0))))) → B1.1(b.1(b.1(y_0)))
B1.1(b.1(b.0(x))) → A1.0(x)
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(x0))
A1.0(a.0(a.1(b.1(b.0(y_0))))) → B1.1(b.1(b.0(y_0)))
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(a.0(x0)))
B1.1(b.1(b.1(x))) → A1.0(x)
The TRS R consists of the following rules:
b.1(b.1(b.0(x))) → a.0(x)
a.0(a.0(A.0(x))) → a.1(b.0(a.0(B.0(x))))
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.0(B.0(x))))
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.0(B.1(x))))
b.1(x0) → b.0(x0)
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.1(b.0(A.1(x)))))
a.0(a.1(x)) → a.1(b.0(a.1(x)))
a.0(a.0(a.0(a.0(A.1(x))))) → a.0(a.1(b.0(a.0(B.1(x)))))
a.0(a.0(a.0(a.0(A.0(x))))) → a.0(a.1(b.0(a.0(B.0(x)))))
a.1(x0) → a.0(x0)
b.1(b.1(b.1(x))) → a.1(x)
a.0(a.0(x)) → a.1(b.0(a.0(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.1(b.0(A.1(x)))))
a.0(a.0(A.0(x))) → a.1(b.0(a.1(b.0(A.0(x)))))
a.0(a.0(a.0(A.0(x)))) → a.1(b.0(a.0(a.0(B.0(x)))))
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.1(b.0(A.0(x)))))
a.0(a.0(a.0(x))) → a.0(a.1(b.0(x)))
b.1(b.0(B.0(x))) → A.0(x)
a.0(a.0(a.1(x))) → a.0(a.1(b.1(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.0(B.1(x))))
a.0(a.0(a.0(A.1(x)))) → a.1(b.0(a.0(a.0(B.1(x)))))
b.1(b.0(B.1(x))) → A.1(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:
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.0(a.1(x0)))
A1.0(a.0(a.1(b.1(b.1(x0))))) → A1.0(a.1(x0))
A1.0(a.0(a.1(b.1(b.1(y_0))))) → B1.1(b.1(b.1(y_0)))
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(x0))
A1.0(a.0(a.1(b.1(b.0(y_0))))) → B1.1(b.1(b.0(y_0)))
A1.0(a.0(a.1(b.1(b.0(x0))))) → A1.0(a.0(a.0(x0)))
B1.1(b.1(b.1(x))) → A1.0(x)
Strictly oriented rules of the TRS R:
b.1(b.1(b.0(x))) → a.0(x)
a.0(a.0(A.0(x))) → a.1(b.0(a.0(B.0(x))))
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.0(B.0(x))))
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.0(B.1(x))))
b.1(x0) → b.0(x0)
a.0(a.0(a.0(a.0(A.1(x))))) → a.0(a.1(b.0(a.0(B.1(x)))))
a.0(a.0(a.0(a.0(A.0(x))))) → a.0(a.1(b.0(a.0(B.0(x)))))
b.1(b.1(b.1(x))) → a.1(x)
a.0(a.0(a.0(A.0(x)))) → a.1(b.0(a.0(a.0(B.0(x)))))
a.0(a.0(a.0(x))) → a.0(a.1(b.0(x)))
b.1(b.0(B.1(x))) → A.1(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = 1 + x1
POL(A.1(x1)) = x1
POL(A1.0(x1)) = 1 + x1
POL(B.0(x1)) = x1
POL(B.1(x1)) = x1
POL(B1.1(x1)) = x1
POL(a.0(x1)) = 1 + x1
POL(a.1(x1)) = 1 + x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = 1 + x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1.1(b.1(b.0(x))) → A1.0(x)
The TRS R consists of the following rules:
a.0(a.0(a.0(A.1(x)))) → a.0(a.1(b.1(b.0(A.1(x)))))
a.0(a.1(x)) → a.1(b.0(a.1(x)))
a.1(x0) → a.0(x0)
a.0(a.0(x)) → a.1(b.0(a.0(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.1(b.0(A.1(x)))))
a.0(a.0(A.0(x))) → a.1(b.0(a.1(b.0(A.0(x)))))
a.0(a.0(a.0(A.0(x)))) → a.0(a.1(b.1(b.0(A.0(x)))))
b.1(b.0(B.0(x))) → A.0(x)
a.0(a.0(a.1(x))) → a.0(a.1(b.1(x)))
a.0(a.0(A.1(x))) → a.1(b.0(a.0(B.1(x))))
a.0(a.0(a.0(A.1(x)))) → a.1(b.0(a.0(a.0(B.1(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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(x)
C(a(a(A(x)))) → C(a(B(x)))
C(a(A(x))) → C(b(A(x)))
C(b(x)) → C(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
c(a(A(x))) → b(b(c(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(a(b(B(x))))
c(a(a(A(x)))) → b(b(c(a(B(x)))))
b(b(B(x))) → A(x)
a(a(a(A(x)))) → a(b(a(a(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(x)
C(a(a(A(x)))) → C(a(B(x)))
C(a(A(x))) → C(b(A(x)))
C(b(x)) → C(x)
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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(x)
C(a(a(A(x)))) → C(a(B(x)))
C(a(A(x))) → C(b(A(x)))
C(b(x)) → C(x)
R is empty.
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:
C(a(x)) → C(x)
C(a(a(A(x)))) → C(a(B(x)))
C(a(A(x))) → C(b(A(x)))
C(b(x)) → C(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 1 + 2·x1
POL(B(x1)) = 1 + x1
POL(C(x1)) = 2·x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = 2 + x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ 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.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
c(a(A(x))) → b(b(c(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(a(b(B(x))))
c(a(a(A(x)))) → b(b(c(a(B(x)))))
b(b(B(x))) → A(x)
a(a(a(A(x)))) → a(b(a(a(B(x)))))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(a(x)))
A(a(a(a(x)))) → A(b(b(a(a(x)))))
A(a(c(x))) → A(b(c(b(b(x)))))
A(a(a(x))) → B(a(b(a(x))))
A(a(a(x))) → A(b(a(b(a(x)))))
A(a(a(a(a(x))))) → B(a(b(a(a(x)))))
A(a(a(a(x)))) → B(b(a(a(x))))
A(a(a(c(x)))) → B(a(c(b(b(x)))))
B(b(b(x))) → A(x)
A(a(a(a(x)))) → B(a(a(b(a(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(a(x)))
A(a(a(a(x)))) → A(b(b(a(a(x)))))
A(a(c(x))) → A(b(c(b(b(x)))))
A(a(a(x))) → B(a(b(a(x))))
A(a(a(x))) → A(b(a(b(a(x)))))
A(a(a(a(a(x))))) → B(a(b(a(a(x)))))
A(a(a(a(x)))) → B(b(a(a(x))))
A(a(a(c(x)))) → B(a(c(b(b(x)))))
B(b(b(x))) → A(x)
A(a(a(a(x)))) → B(a(a(b(a(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
a(a(a(A(x)))) → a(a(b(b(A(x)))))
c(a(A(x))) → b(b(c(b(A(x)))))
a(a(A(x))) → a(b(a(B(x))))
a(a(A(x))) → a(b(a(b(A(x)))))
a(a(a(a(A(x))))) → a(a(b(a(B(x)))))
a(a(a(A(x)))) → a(a(b(B(x))))
c(a(a(A(x)))) → b(b(c(a(B(x)))))
b(b(B(x))) → A(x)
a(a(a(A(x)))) → a(b(a(a(B(x)))))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(a(x)))
A(a(a(a(x)))) → A(b(b(a(a(x)))))
A(a(c(x))) → A(b(c(b(b(x)))))
A(a(a(x))) → B(a(b(a(x))))
A(a(a(x))) → A(b(a(b(a(x)))))
A(a(a(a(a(x))))) → B(a(b(a(a(x)))))
A(a(a(a(x)))) → B(b(a(a(x))))
A(a(a(c(x)))) → B(a(c(b(b(x)))))
B(b(b(x))) → A(x)
A(a(a(a(x)))) → B(a(a(b(a(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
b(c(x)) → c(a(a(x)))
a(c(x)) → c(b(b(x)))
a(a(a(x))) → b(a(a(x)))
A(a(a(a(x)))) → A(b(b(a(a(x)))))
A(a(c(x))) → A(b(c(b(b(x)))))
A(a(a(x))) → B(a(b(a(x))))
A(a(a(x))) → A(b(a(b(a(x)))))
A(a(a(a(a(x))))) → B(a(b(a(a(x)))))
A(a(a(a(x)))) → B(b(a(a(x))))
A(a(a(c(x)))) → B(a(c(b(b(x)))))
B(b(b(x))) → A(x)
A(a(a(a(x)))) → B(a(a(b(a(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(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(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x1))) → a(x1)
a(a(x1)) → a(b(a(x1)))
b(c(x1)) → c(a(a(x1)))
a(c(x1)) → c(b(b(x1)))
a(a(a(x1))) → b(a(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(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(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
c(b(x)) → a(a(c(x)))
c(a(x)) → b(b(c(x)))
a(a(a(x))) → a(a(b(x)))
Q is empty.