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(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(sel2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(g1(X)) -> ACTIVE1(X)
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(f1(X)) -> G1(X)
ACTIVE1(g1(s1(X))) -> G1(X)
ACTIVE1(g1(s1(X))) -> S1(g1(X))
PROPER1(sel2(X1, X2)) -> SEL2(proper1(X1), proper1(X2))
S1(mark1(X)) -> S1(X)
G1(mark1(X)) -> G1(X)
ACTIVE1(sel2(X1, X2)) -> SEL2(X1, active1(X2))
PROPER1(f1(X)) -> PROPER1(X)
ACTIVE1(sel2(s1(X), cons2(Y, Z))) -> SEL2(X, Z)
SEL2(X1, mark1(X2)) -> SEL2(X1, X2)
ACTIVE1(sel2(X1, X2)) -> ACTIVE1(X1)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
PROPER1(sel2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
SEL2(ok1(X1), ok1(X2)) -> SEL2(X1, X2)
ACTIVE1(f1(X)) -> F1(active1(X))
ACTIVE1(g1(s1(X))) -> S1(s1(g1(X)))
PROPER1(sel2(X1, X2)) -> PROPER1(X2)
TOP1(ok1(X)) -> ACTIVE1(X)
SEL2(mark1(X1), X2) -> SEL2(X1, X2)
F1(mark1(X)) -> F1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
ACTIVE1(g1(X)) -> G1(active1(X))
ACTIVE1(f1(X)) -> F1(g1(X))
TOP1(ok1(X)) -> TOP1(active1(X))
F1(ok1(X)) -> F1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
ACTIVE1(sel2(X1, X2)) -> SEL2(active1(X1), X2)
ACTIVE1(g1(0)) -> S1(0)
G1(ok1(X)) -> G1(X)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(f1(X)) -> CONS2(X, f1(g1(X)))
PROPER1(f1(X)) -> F1(proper1(X))
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(sel2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(g1(X)) -> ACTIVE1(X)
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(f1(X)) -> G1(X)
ACTIVE1(g1(s1(X))) -> G1(X)
ACTIVE1(g1(s1(X))) -> S1(g1(X))
PROPER1(sel2(X1, X2)) -> SEL2(proper1(X1), proper1(X2))
S1(mark1(X)) -> S1(X)
G1(mark1(X)) -> G1(X)
ACTIVE1(sel2(X1, X2)) -> SEL2(X1, active1(X2))
PROPER1(f1(X)) -> PROPER1(X)
ACTIVE1(sel2(s1(X), cons2(Y, Z))) -> SEL2(X, Z)
SEL2(X1, mark1(X2)) -> SEL2(X1, X2)
ACTIVE1(sel2(X1, X2)) -> ACTIVE1(X1)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
PROPER1(sel2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
SEL2(ok1(X1), ok1(X2)) -> SEL2(X1, X2)
ACTIVE1(f1(X)) -> F1(active1(X))
ACTIVE1(g1(s1(X))) -> S1(s1(g1(X)))
PROPER1(sel2(X1, X2)) -> PROPER1(X2)
TOP1(ok1(X)) -> ACTIVE1(X)
SEL2(mark1(X1), X2) -> SEL2(X1, X2)
F1(mark1(X)) -> F1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
ACTIVE1(g1(X)) -> G1(active1(X))
ACTIVE1(f1(X)) -> F1(g1(X))
TOP1(ok1(X)) -> TOP1(active1(X))
F1(ok1(X)) -> F1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
ACTIVE1(sel2(X1, X2)) -> SEL2(active1(X1), X2)
ACTIVE1(g1(0)) -> S1(0)
G1(ok1(X)) -> G1(X)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(f1(X)) -> CONS2(X, f1(g1(X)))
PROPER1(f1(X)) -> F1(proper1(X))
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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 8 SCCs with 21 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL2(X1, mark1(X2)) -> SEL2(X1, X2)
SEL2(ok1(X1), ok1(X2)) -> SEL2(X1, X2)
SEL2(mark1(X1), X2) -> SEL2(X1, X2)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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.
SEL2(ok1(X1), ok1(X2)) -> SEL2(X1, X2)
Used argument filtering: SEL2(x1, x2) = x2
mark1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL2(X1, mark1(X2)) -> SEL2(X1, X2)
SEL2(mark1(X1), X2) -> SEL2(X1, X2)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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.
SEL2(X1, mark1(X2)) -> SEL2(X1, X2)
Used argument filtering: SEL2(x1, x2) = x2
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL2(mark1(X1), X2) -> SEL2(X1, X2)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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.
SEL2(mark1(X1), X2) -> SEL2(X1, X2)
Used argument filtering: SEL2(x1, x2) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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.
S1(mark1(X)) -> S1(X)
Used argument filtering: S1(x1) = x1
ok1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S1(ok1(X)) -> S1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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.
S1(ok1(X)) -> S1(X)
Used argument filtering: S1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(mark1(X)) -> G1(X)
G1(ok1(X)) -> G1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
mark1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(mark1(X)) -> G1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(mark1(X)) -> G1(X)
Used argument filtering: G1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ 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(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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.
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2) = x2
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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.
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ 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(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ QDP
↳ QDPAfsSolverProof
↳ 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(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
mark1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ 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(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(mark1(X)) -> F1(X)
Used argument filtering: F1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ 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(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
PROPER1(sel2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(sel2(X1, X2)) -> PROPER1(X2)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(sel2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(sel2(X1, X2)) -> PROPER1(X2)
Used argument filtering: PROPER1(x1) = x1
g1(x1) = x1
s1(x1) = x1
f1(x1) = x1
sel2(x1, x2) = sel2(x1, x2)
cons2(x1, x2) = cons2(x1, x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(s1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
g1(x1) = x1
f1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ 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(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ 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(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(sel2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(g1(X)) -> ACTIVE1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(sel2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(sel2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(sel2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1) = x1
sel2(x1, x2) = sel2(x1, x2)
g1(x1) = x1
f1(x1) = x1
cons2(x1, x2) = x1
s1(x1) = x1
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(g1(X)) -> ACTIVE1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(s1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
g1(x1) = x1
f1(x1) = x1
cons2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(g1(X)) -> ACTIVE1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(cons2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1) = x1
g1(x1) = x1
f1(x1) = x1
cons2(x1, x2) = cons1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(g1(X)) -> ACTIVE1(X)
ACTIVE1(f1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(f1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
g1(x1) = x1
f1(x1) = f1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(g1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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(g1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
g1(x1) = g1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ 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:
active1(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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
↳ 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(f1(X)) -> mark1(cons2(X, f1(g1(X))))
active1(g1(0)) -> mark1(s1(0))
active1(g1(s1(X))) -> mark1(s1(s1(g1(X))))
active1(sel2(0, cons2(X, Y))) -> mark1(X)
active1(sel2(s1(X), cons2(Y, Z))) -> mark1(sel2(X, Z))
active1(f1(X)) -> f1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(g1(X)) -> g1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(sel2(X1, X2)) -> sel2(active1(X1), X2)
active1(sel2(X1, X2)) -> sel2(X1, active1(X2))
f1(mark1(X)) -> mark1(f1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
g1(mark1(X)) -> mark1(g1(X))
s1(mark1(X)) -> mark1(s1(X))
sel2(mark1(X1), X2) -> mark1(sel2(X1, X2))
sel2(X1, mark1(X2)) -> mark1(sel2(X1, X2))
proper1(f1(X)) -> f1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(g1(X)) -> g1(proper1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sel2(X1, X2)) -> sel2(proper1(X1), proper1(X2))
f1(ok1(X)) -> ok1(f1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
g1(ok1(X)) -> ok1(g1(X))
s1(ok1(X)) -> ok1(s1(X))
sel2(ok1(X1), ok1(X2)) -> ok1(sel2(X1, X2))
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.