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:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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:
ACTIVE1(f3(x, y, z)) -> ACTIVE1(z)
PROPER1(f3(x, y, z)) -> PROPER1(x)
TOP1(mark1(x)) -> PROPER1(x)
PROPER1(f3(x, y, z)) -> PROPER1(z)
TOP1(ok1(x)) -> ACTIVE1(x)
ACTIVE1(f3(x, y, z)) -> F3(x, y, active1(z))
ACTIVE1(f3(b, c, x)) -> F3(x, x, x)
TOP1(ok1(x)) -> TOP1(active1(x))
TOP1(mark1(x)) -> TOP1(proper1(x))
PROPER1(f3(x, y, z)) -> PROPER1(y)
PROPER1(f3(x, y, z)) -> F3(proper1(x), proper1(y), proper1(z))
F3(x, y, mark1(z)) -> F3(x, y, z)
F3(ok1(x), ok1(y), ok1(z)) -> F3(x, y, z)
The TRS R consists of the following rules:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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:
ACTIVE1(f3(x, y, z)) -> ACTIVE1(z)
PROPER1(f3(x, y, z)) -> PROPER1(x)
TOP1(mark1(x)) -> PROPER1(x)
PROPER1(f3(x, y, z)) -> PROPER1(z)
TOP1(ok1(x)) -> ACTIVE1(x)
ACTIVE1(f3(x, y, z)) -> F3(x, y, active1(z))
ACTIVE1(f3(b, c, x)) -> F3(x, x, x)
TOP1(ok1(x)) -> TOP1(active1(x))
TOP1(mark1(x)) -> TOP1(proper1(x))
PROPER1(f3(x, y, z)) -> PROPER1(y)
PROPER1(f3(x, y, z)) -> F3(proper1(x), proper1(y), proper1(z))
F3(x, y, mark1(z)) -> F3(x, y, z)
F3(ok1(x), ok1(y), ok1(z)) -> F3(x, y, z)
The TRS R consists of the following rules:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F3(x, y, mark1(z)) -> F3(x, y, z)
F3(ok1(x), ok1(y), ok1(z)) -> F3(x, y, z)
The TRS R consists of the following rules:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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.
F3(ok1(x), ok1(y), ok1(z)) -> F3(x, y, z)
Used argument filtering: F3(x1, x2, x3) = x3
mark1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F3(x, y, mark1(z)) -> F3(x, y, z)
The TRS R consists of the following rules:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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.
F3(x, y, mark1(z)) -> F3(x, y, z)
Used argument filtering: F3(x1, x2, x3) = x3
mark1(x1) = mark1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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:
PROPER1(f3(x, y, z)) -> PROPER1(x)
PROPER1(f3(x, y, z)) -> PROPER1(z)
PROPER1(f3(x, y, z)) -> PROPER1(y)
The TRS R consists of the following rules:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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(f3(x, y, z)) -> PROPER1(x)
PROPER1(f3(x, y, z)) -> PROPER1(z)
PROPER1(f3(x, y, z)) -> PROPER1(y)
Used argument filtering: PROPER1(x1) = x1
f3(x1, x2, x3) = f3(x1, x2, x3)
Used ordering: Quasi 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(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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:
ACTIVE1(f3(x, y, z)) -> ACTIVE1(z)
The TRS R consists of the following rules:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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.
ACTIVE1(f3(x, y, z)) -> ACTIVE1(z)
Used argument filtering: ACTIVE1(x1) = x1
f3(x1, x2, x3) = f1(x3)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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
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(f3(b, c, x)) -> mark1(f3(x, x, x))
active1(f3(x, y, z)) -> f3(x, y, active1(z))
active1(d) -> m1(b)
f3(x, y, mark1(z)) -> mark1(f3(x, y, z))
active1(d) -> mark1(c)
proper1(b) -> ok1(b)
proper1(c) -> ok1(c)
proper1(d) -> ok1(d)
proper1(f3(x, y, z)) -> f3(proper1(x), proper1(y), proper1(z))
f3(ok1(x), ok1(y), ok1(z)) -> ok1(f3(x, y, z))
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.