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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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)
ACTIVE1(zeros) -> CONS2(0, zeros)
ACTIVE1(adx1(cons2(X, Y))) -> CONS2(X, adx1(Y))
ACTIVE1(tl1(X)) -> TL1(active1(X))
ACTIVE1(incr1(cons2(X, Y))) -> INCR1(Y)
ACTIVE1(tl1(X)) -> ACTIVE1(X)
PROPER1(adx1(X)) -> PROPER1(X)
INCR1(mark1(X)) -> INCR1(X)
TL1(ok1(X)) -> TL1(X)
PROPER1(adx1(X)) -> ADX1(proper1(X))
ACTIVE1(adx1(cons2(X, Y))) -> ADX1(Y)
PROPER1(incr1(X)) -> INCR1(proper1(X))
PROPER1(incr1(X)) -> PROPER1(X)
ACTIVE1(incr1(cons2(X, Y))) -> S1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(hd1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
ACTIVE1(nats) -> ADX1(zeros)
PROPER1(hd1(X)) -> HD1(proper1(X))
INCR1(ok1(X)) -> INCR1(X)
PROPER1(hd1(X)) -> PROPER1(X)
ADX1(mark1(X)) -> ADX1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(adx1(cons2(X, Y))) -> INCR1(cons2(X, adx1(Y)))
ACTIVE1(incr1(X)) -> INCR1(active1(X))
ACTIVE1(incr1(cons2(X, Y))) -> CONS2(s1(X), incr1(Y))
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
PROPER1(tl1(X)) -> TL1(proper1(X))
HD1(mark1(X)) -> HD1(X)
ADX1(ok1(X)) -> ADX1(X)
ACTIVE1(hd1(X)) -> HD1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
HD1(ok1(X)) -> HD1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(adx1(X)) -> ADX1(active1(X))
TL1(mark1(X)) -> TL1(X)
PROPER1(tl1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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)
ACTIVE1(zeros) -> CONS2(0, zeros)
ACTIVE1(adx1(cons2(X, Y))) -> CONS2(X, adx1(Y))
ACTIVE1(tl1(X)) -> TL1(active1(X))
ACTIVE1(incr1(cons2(X, Y))) -> INCR1(Y)
ACTIVE1(tl1(X)) -> ACTIVE1(X)
PROPER1(adx1(X)) -> PROPER1(X)
INCR1(mark1(X)) -> INCR1(X)
TL1(ok1(X)) -> TL1(X)
PROPER1(adx1(X)) -> ADX1(proper1(X))
ACTIVE1(adx1(cons2(X, Y))) -> ADX1(Y)
PROPER1(incr1(X)) -> INCR1(proper1(X))
PROPER1(incr1(X)) -> PROPER1(X)
ACTIVE1(incr1(cons2(X, Y))) -> S1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(hd1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
ACTIVE1(nats) -> ADX1(zeros)
PROPER1(hd1(X)) -> HD1(proper1(X))
INCR1(ok1(X)) -> INCR1(X)
PROPER1(hd1(X)) -> PROPER1(X)
ADX1(mark1(X)) -> ADX1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(adx1(cons2(X, Y))) -> INCR1(cons2(X, adx1(Y)))
ACTIVE1(incr1(X)) -> INCR1(active1(X))
ACTIVE1(incr1(cons2(X, Y))) -> CONS2(s1(X), incr1(Y))
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
PROPER1(tl1(X)) -> TL1(proper1(X))
HD1(mark1(X)) -> HD1(X)
ADX1(ok1(X)) -> ADX1(X)
ACTIVE1(hd1(X)) -> HD1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
HD1(ok1(X)) -> HD1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(adx1(X)) -> ADX1(active1(X))
TL1(mark1(X)) -> TL1(X)
PROPER1(tl1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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 9 SCCs with 20 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ 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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
S1(ok1(X)) -> S1(X)
Used argument filtering: S1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2) = x2
ok1(x1) = ok1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TL1(ok1(X)) -> TL1(X)
TL1(mark1(X)) -> TL1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
TL1(mark1(X)) -> TL1(X)
Used argument filtering: TL1(x1) = x1
ok1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TL1(ok1(X)) -> TL1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
TL1(ok1(X)) -> TL1(X)
Used argument filtering: TL1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HD1(mark1(X)) -> HD1(X)
HD1(ok1(X)) -> HD1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
HD1(ok1(X)) -> HD1(X)
Used argument filtering: HD1(x1) = x1
mark1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HD1(mark1(X)) -> HD1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
HD1(mark1(X)) -> HD1(X)
Used argument filtering: HD1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INCR1(ok1(X)) -> INCR1(X)
INCR1(mark1(X)) -> INCR1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
INCR1(mark1(X)) -> INCR1(X)
Used argument filtering: INCR1(x1) = x1
ok1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INCR1(ok1(X)) -> INCR1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
INCR1(ok1(X)) -> INCR1(X)
Used argument filtering: INCR1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ADX1(ok1(X)) -> ADX1(X)
ADX1(mark1(X)) -> ADX1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
ADX1(mark1(X)) -> ADX1(X)
Used argument filtering: ADX1(x1) = x1
ok1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ADX1(ok1(X)) -> ADX1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
ADX1(ok1(X)) -> ADX1(X)
Used argument filtering: ADX1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(hd1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
Used argument filtering: PROPER1(x1) = x1
hd1(x1) = x1
s1(x1) = x1
cons2(x1, x2) = cons2(x1, x2)
incr1(x1) = x1
adx1(x1) = x1
tl1(x1) = x1
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(hd1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(adx1(X)) -> PROPER1(X)
PROPER1(tl1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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(tl1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
s1(x1) = x1
hd1(x1) = x1
incr1(x1) = x1
adx1(x1) = x1
tl1(x1) = tl1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(hd1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(adx1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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(adx1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
hd1(x1) = x1
s1(x1) = x1
incr1(x1) = x1
adx1(x1) = adx1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(s1(X)) -> PROPER1(X)
PROPER1(hd1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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(incr1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
s1(x1) = x1
hd1(x1) = x1
incr1(x1) = incr1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(hd1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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(s1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
hd1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(hd1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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(hd1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
hd1(x1) = hd1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ 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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(hd1(X)) -> ACTIVE1(X)
ACTIVE1(tl1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
ACTIVE1(incr1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
hd1(x1) = x1
tl1(x1) = x1
adx1(x1) = x1
incr1(x1) = incr1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(hd1(X)) -> ACTIVE1(X)
ACTIVE1(tl1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
ACTIVE1(adx1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
hd1(x1) = x1
tl1(x1) = x1
adx1(x1) = adx1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(hd1(X)) -> ACTIVE1(X)
ACTIVE1(tl1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
ACTIVE1(tl1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
hd1(x1) = x1
tl1(x1) = tl1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(hd1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.
ACTIVE1(hd1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
hd1(x1) = hd1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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
↳ 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(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(incr1(cons2(X, Y))) -> mark1(cons2(s1(X), incr1(Y)))
active1(adx1(cons2(X, Y))) -> mark1(incr1(cons2(X, adx1(Y))))
active1(hd1(cons2(X, Y))) -> mark1(X)
active1(tl1(cons2(X, Y))) -> mark1(Y)
active1(adx1(X)) -> adx1(active1(X))
active1(incr1(X)) -> incr1(active1(X))
active1(hd1(X)) -> hd1(active1(X))
active1(tl1(X)) -> tl1(active1(X))
adx1(mark1(X)) -> mark1(adx1(X))
incr1(mark1(X)) -> mark1(incr1(X))
hd1(mark1(X)) -> mark1(hd1(X))
tl1(mark1(X)) -> mark1(tl1(X))
proper1(nats) -> ok1(nats)
proper1(adx1(X)) -> adx1(proper1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(incr1(X)) -> incr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(hd1(X)) -> hd1(proper1(X))
proper1(tl1(X)) -> tl1(proper1(X))
adx1(ok1(X)) -> ok1(adx1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
incr1(ok1(X)) -> ok1(incr1(X))
s1(ok1(X)) -> ok1(s1(X))
hd1(ok1(X)) -> ok1(hd1(X))
tl1(ok1(X)) -> ok1(tl1(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.