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:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(c) -> F1(g1(c))
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
ACTIVE1(c) -> G1(c)
TOP1(ok1(X)) -> ACTIVE1(X)
PROPER1(f1(X)) -> F1(proper1(X))
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(c) -> F1(g1(c))
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
ACTIVE1(c) -> G1(c)
TOP1(ok1(X)) -> ACTIVE1(X)
PROPER1(f1(X)) -> F1(proper1(X))
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(ok1(X)) -> G1(X)
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.
G1(ok1(X)) -> G1(X)
Used argument filtering: G1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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
Q DP problem:
The TRS P consists of the following rules:
F1(ok1(X)) -> F1(X)
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.
F1(ok1(X)) -> F1(X)
Used argument filtering: F1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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
Q DP problem:
The TRS P consists of the following rules:
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.
PROPER1(f1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
g1(x1) = x1
f1(x1) = f1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(g1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.
PROPER1(g1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
g1(x1) = g1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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
Q DP problem:
The TRS P consists of the following rules:
TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(mark1(X)) -> TOP1(proper1(X))
Used argument filtering: TOP1(x1) = x1
ok1(x1) = x1
active1(x1) = x1
mark1(x1) = mark1(x1)
proper1(x1) = x1
c = c
f1(x1) = f
g1(x1) = g
Used ordering: Precedence:
c > f > mark1
c > f > g
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
TOP1(ok1(X)) -> TOP1(active1(X))
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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(ok1(X)) -> TOP1(active1(X))
Used argument filtering: TOP1(x1) = x1
ok1(x1) = ok
active1(x1) = active
mark1(x1) = mark
g1(x1) = g
Used ordering: Precedence:
g > ok > active > mark
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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.