Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
TOP1(rec1(up1(x))) -> REC1(x)
CHECK1(no1(x)) -> CHECK1(x)
REC1(bot) -> SENT1(bot)
REC1(rec1(x)) -> SENT1(rec1(x))
TOP1(sent1(up1(x))) -> CHECK1(rec1(x))
CHECK1(rec1(x)) -> CHECK1(x)
REC1(up1(x)) -> REC1(x)
TOP1(no1(up1(x))) -> REC1(x)
TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(no1(x)) -> SENT1(rec1(x))
CHECK1(sent1(x)) -> SENT1(check1(x))
CHECK1(rec1(x)) -> REC1(check1(x))
TOP1(no1(up1(x))) -> CHECK1(rec1(x))
CHECK1(no1(x)) -> NO1(check1(x))
REC1(sent1(x)) -> SENT1(rec1(x))
CHECK1(up1(x)) -> CHECK1(x)
TOP1(sent1(up1(x))) -> REC1(x)
TOP1(rec1(up1(x))) -> CHECK1(rec1(x))
TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(no1(x)) -> REC1(x)
NO1(up1(x)) -> NO1(x)
SENT1(up1(x)) -> SENT1(x)
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(sent1(x)) -> REC1(x)
CHECK1(sent1(x)) -> CHECK1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
TOP1(rec1(up1(x))) -> REC1(x)
CHECK1(no1(x)) -> CHECK1(x)
REC1(bot) -> SENT1(bot)
REC1(rec1(x)) -> SENT1(rec1(x))
TOP1(sent1(up1(x))) -> CHECK1(rec1(x))
CHECK1(rec1(x)) -> CHECK1(x)
REC1(up1(x)) -> REC1(x)
TOP1(no1(up1(x))) -> REC1(x)
TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(no1(x)) -> SENT1(rec1(x))
CHECK1(sent1(x)) -> SENT1(check1(x))
CHECK1(rec1(x)) -> REC1(check1(x))
TOP1(no1(up1(x))) -> CHECK1(rec1(x))
CHECK1(no1(x)) -> NO1(check1(x))
REC1(sent1(x)) -> SENT1(rec1(x))
CHECK1(up1(x)) -> CHECK1(x)
TOP1(sent1(up1(x))) -> REC1(x)
TOP1(rec1(up1(x))) -> CHECK1(rec1(x))
TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(no1(x)) -> REC1(x)
NO1(up1(x)) -> NO1(x)
SENT1(up1(x)) -> SENT1(x)
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(sent1(x)) -> REC1(x)
CHECK1(sent1(x)) -> CHECK1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 13 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
NO1(up1(x)) -> NO1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
NO1(up1(x)) -> NO1(x)
Used argument filtering: NO1(x1) = x1
up1(x1) = up1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
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
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SENT1(up1(x)) -> SENT1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
SENT1(up1(x)) -> SENT1(x)
Used argument filtering: SENT1(x1) = x1
up1(x1) = up1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
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
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
REC1(up1(x)) -> REC1(x)
REC1(no1(x)) -> REC1(x)
REC1(sent1(x)) -> REC1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
REC1(sent1(x)) -> REC1(x)
Used argument filtering: REC1(x1) = x1
up1(x1) = x1
no1(x1) = x1
sent1(x1) = sent1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
REC1(up1(x)) -> REC1(x)
REC1(no1(x)) -> REC1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
REC1(no1(x)) -> REC1(x)
Used argument filtering: REC1(x1) = x1
up1(x1) = x1
no1(x1) = no1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
REC1(up1(x)) -> REC1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
REC1(up1(x)) -> REC1(x)
Used argument filtering: REC1(x1) = x1
up1(x1) = up1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
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
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CHECK1(no1(x)) -> CHECK1(x)
CHECK1(up1(x)) -> CHECK1(x)
CHECK1(sent1(x)) -> CHECK1(x)
CHECK1(rec1(x)) -> CHECK1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
CHECK1(rec1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1) = x1
no1(x1) = x1
up1(x1) = x1
sent1(x1) = x1
rec1(x1) = rec1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CHECK1(no1(x)) -> CHECK1(x)
CHECK1(up1(x)) -> CHECK1(x)
CHECK1(sent1(x)) -> CHECK1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
CHECK1(sent1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1) = x1
no1(x1) = x1
up1(x1) = x1
sent1(x1) = sent1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CHECK1(no1(x)) -> CHECK1(x)
CHECK1(up1(x)) -> CHECK1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
CHECK1(up1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1) = x1
no1(x1) = x1
up1(x1) = up1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CHECK1(no1(x)) -> CHECK1(x)
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
CHECK1(no1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1) = x1
no1(x1) = no1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
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
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
Used argument filtering: TOP1(x1) = x1
no1(x1) = no1(x1)
up1(x1) = x1
check1(x1) = x1
rec1(x1) = x1
sent1(x1) = x1
bot = bot
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
The TRS R consists of the following rules:
rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.