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(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
TOP1(found1(x)) -> TOP1(active1(x))
CHECK1(f1(x)) -> CHECK1(x)
MATCH2(X, x) -> PROPER1(x)
ACTIVE1(f1(x)) -> ACTIVE1(x)
TOP1(mark1(x)) -> CHECK1(x)
ACTIVE1(f1(x)) -> F1(active1(x))
CHECK1(x) -> MATCH2(f1(X), x)
TOP1(mark1(x)) -> TOP1(check1(x))
CHECK1(f1(x)) -> F1(check1(x))
F1(mark1(x)) -> F1(x)
CHECK1(x) -> F1(X)
MATCH2(f1(x), f1(y)) -> F1(match2(x, y))
PROPER1(f1(x)) -> PROPER1(x)
TOP1(found1(x)) -> ACTIVE1(x)
F1(ok1(x)) -> F1(x)
CHECK1(x) -> START1(match2(f1(X), x))
TOP1(active1(c)) -> TOP1(mark1(c))
MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
F1(found1(x)) -> F1(x)
PROPER1(f1(x)) -> F1(proper1(x))
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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(found1(x)) -> TOP1(active1(x))
CHECK1(f1(x)) -> CHECK1(x)
MATCH2(X, x) -> PROPER1(x)
ACTIVE1(f1(x)) -> ACTIVE1(x)
TOP1(mark1(x)) -> CHECK1(x)
ACTIVE1(f1(x)) -> F1(active1(x))
CHECK1(x) -> MATCH2(f1(X), x)
TOP1(mark1(x)) -> TOP1(check1(x))
CHECK1(f1(x)) -> F1(check1(x))
F1(mark1(x)) -> F1(x)
CHECK1(x) -> F1(X)
MATCH2(f1(x), f1(y)) -> F1(match2(x, y))
PROPER1(f1(x)) -> PROPER1(x)
TOP1(found1(x)) -> ACTIVE1(x)
F1(ok1(x)) -> F1(x)
CHECK1(x) -> START1(match2(f1(X), x))
TOP1(active1(c)) -> TOP1(mark1(c))
MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
F1(found1(x)) -> F1(x)
PROPER1(f1(x)) -> F1(proper1(x))
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 6 SCCs with 10 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(mark1(x)) -> F1(x)
F1(ok1(x)) -> F1(x)
F1(found1(x)) -> F1(x)
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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(found1(x)) -> F1(x)
Used argument filtering: F1(x1) = x1
mark1(x1) = x1
ok1(x1) = x1
found1(x1) = found1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(mark1(x)) -> F1(x)
F1(ok1(x)) -> F1(x)
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
mark1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(mark1(x)) -> F1(x)
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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(mark1(x)) -> F1(x)
Used argument filtering: F1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(f1(x)) -> ACTIVE1(x)
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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(f1(x)) -> ACTIVE1(x)
Used argument filtering: ACTIVE1(x1) = x1
f1(x1) = f1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(f1(x)) -> PROPER1(x)
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
f1(x1) = f1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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.
MATCH2(f1(x), f1(y)) -> MATCH2(x, y)
Used argument filtering: MATCH2(x1, x2) = x2
f1(x1) = f1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CHECK1(f1(x)) -> CHECK1(x)
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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(f1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1) = x1
f1(x1) = f1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
TOP1(found1(x)) -> TOP1(active1(x))
TOP1(active1(c)) -> TOP1(mark1(c))
TOP1(mark1(x)) -> TOP1(check1(x))
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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(active1(c)) -> TOP1(mark1(c))
Used argument filtering: TOP1(x1) = x1
found1(x1) = x1
active1(x1) = x1
c = c
mark1(x1) = mark
check1(x1) = check
f1(x1) = f
ok1(x1) = x1
start1(x1) = x1
match2(x1, x2) = match
Used ordering: Quasi Precedence:
c > [mark, check, f, match]
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
TOP1(found1(x)) -> TOP1(active1(x))
TOP1(mark1(x)) -> TOP1(check1(x))
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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(check1(x))
Used argument filtering: TOP1(x1) = x1
found1(x1) = x1
active1(x1) = x1
mark1(x1) = mark1(x1)
check1(x1) = x1
f1(x1) = f1(x1)
start1(x1) = x1
match2(x1, x2) = x2
ok1(x1) = x1
proper1(x1) = x1
c = c
Used ordering: Quasi Precedence:
f_1 > mark_1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
TOP1(found1(x)) -> TOP1(active1(x))
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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(found1(x)) -> TOP1(active1(x))
Used argument filtering: TOP1(x1) = x1
found1(x1) = found
active1(x1) = active
mark1(x1) = mark
f1(x1) = x1
ok1(x1) = ok
Used ordering: Quasi Precedence:
found > active > mark
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(x)) -> mark1(x)
top1(active1(c)) -> top1(mark1(c))
top1(mark1(x)) -> top1(check1(x))
check1(f1(x)) -> f1(check1(x))
check1(x) -> start1(match2(f1(X), x))
match2(f1(x), f1(y)) -> f1(match2(x, y))
match2(X, x) -> proper1(x)
proper1(c) -> ok1(c)
proper1(f1(x)) -> f1(proper1(x))
f1(ok1(x)) -> ok1(f1(x))
start1(ok1(x)) -> found1(x)
f1(found1(x)) -> found1(f1(x))
top1(found1(x)) -> top1(active1(x))
active1(f1(x)) -> f1(active1(x))
f1(mark1(x)) -> mark1(f1(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.